·No text on this page.

·No text on this page.

·No text on this page.

·SPHERICAL TRIGONOMETRY.

·OX

·SPHERICAL
TRIGONOMETRY
for the Use of Colleges and Schools.
WITH NUMEROUS EXAMPLES.
I. TODHUNTER, M.A., F.R.S.,
LATE FELLOW AND PRINCIPAL MATHEMATICAL LECTURER OF
ST JOHN'S COLLEGE, CAMBRIDGE.
THIRD EDITION, ENLARGED.
London and New York:
MACMILLAN AND CO.
1871.
[All Rights reserved. )

·Cambridge:
PRINTED BY C. J. ('LAY, M.A.
AT THE UNIVERSITY PRESS.
1. lurpera Jukrishna Iublic Library
loc. N.21.7.22 Danke...!2.9

·PREFACE. The preset 'work is constructed on the same plan as my treatise on Plane Trigonometry, to which it is intended, as a sequel; it contains all the propositions usually included under the head of Spherical Trigonometry, together with a large collection of examples for exercise. In the course of the work reference is made to preceding writers from whom assistance las necn obtained; besides these writers' I have consulte ane treatises on Trigonometry by Lardner, Lefebure de Fourcy, and Showball, and the treatise on Geometry published in the Library of Useful Knowledge. The examples have been Chieriy selected from the University and College Examination Papers. in the account of Napier's Rules of Circular Parts an explanation has been given of a method of proof devised by Napier, which seer'is to have been overlooked by most modern writers on the subject. I have had the advantage of access to an unprivted Memoir on this point by the late R. L. Ellis of Trinity College; Mr Ellis had in fact rediscovered for himself Napier's own method. For the use of this Memoir and for some valuable references on the subject I am indebted to the Dean of Ely. Considerable labour las been bestowed on the text in order to render it comprehensive and accurate, and the examples have all been carefully verified; and thus I venture to hope tilat the work will be found useful by Students and Teachers. 1. TODHUNTER. ST JOHN'S COLLEGE, August 15, 1859.

·• In the third edition I have made some additions which I hope will be found valuable. I have considerably enlargel the discussion on tire connexion of Formule in Planc-au Spherical Trigonometry; so as to include an accomt of the properties in Spherical Trigonometry which are analogous to those of the Nine Points Circle in Plane Geometry. The mode of investigation is inore clementary than those hitherto employed; and perheps some of the results are new. The fourteenth Chapter is almost entirely original, and may deserve attention from the nature ui' the propositious themselves and of the demonstrations which ar given. CAMBRIDGE, July, 1871.

·CONTENTS. (BAT M'AGE I. Great and Small Circles 1 3 II. Spherical Triangles 6g III. Spherical Geometry 10 IV. Relations between the Trigonometrical Functions of the Sides and the Angles of a Splerical Triangle 16 V. Solution of Right-angled Triangles 82 11. Solution of Oblique-angled Triangler 46 VII. Circumscribed and Inscribed Circles 59 III. Arca of a Spherical Triangle. Spherical Excess 67 IX. On ocrtain approximate Formulæ 75 X. Geodetical operations ! 85 XI. On small variatious in the parts of a Spherical Triangle 93 XII. On the connexion of Formulæ in Plane and Spherioal Trigo. nometry 96 XIII. Polyhedrons 114 XIV. Aros drawn to fixed points on the Surface of a Sphere 125 XV. Miscellaneous Propositions 135 XVI. Numerical Solution of Splorical Triangles 149 1

·No text on this page.

·SPHERICAL TRIGONOMETRY. I. GREAT AND SMALL CIRCLES. 1. A SPHERE is a solid bounded by a surface every point of which is equally distant from a fixed point which is called the centre of the sphere. The straight line which joins any point of •the surface with the centre is called a radius. A straight line drawn through the centre and teminated both ways by the surface is called a diameter, 2. The section of the surface of a sphere made by any plant is a circle. B Let AB be the section of the surface of a sphere made by any plme, the centre of the sphere. Draw OC perpendicular to the plane; take any point D in the section and join OD, CD. Since OC is perpendicular to the plans, the angle OCD is a right angle; therefore CD=J(ODS - OC*). Now 0 and C are fixed points, so that OC is constant; and OD is constant, being the radius of the T. S. T. B ·2 GREAT AND SMALL CIRCLES. sphere; hence CD is constant. Thus all points in the plane section are equally distant from the fixed point C; therefore the section is a circle of which C is the centre. 3. The section of the surface of a sphere Ply a plane is calledo a great circle if the plane passes through the centre of the sphere, and a small circle if the plane does not pass through the centre of the sphere. Thus the radius of a great circle is equal to the radius of the sphere. 4. Through the centre of a sphere and any two points on the surface a plane can be drawn; and only one plane can be drawn, except when the two points are the extremities of a diameter of the sphere, and then an infinite number of such planes can be drawn. Hence only one great circle can be drawn through two given points on the surface of a sphere, except when the joints are the extremities of a diameter of the sphere. When only one great circle can be drawn through two given points, the great citele unequally divided at the two points; we shall for brevity speak of the shorter of the two arcs as the arc of a great circle joining the two points. 5. The axis of any circlo of a suhere is 'that diameter of the sphere which is perpendicular to the plane of the circle; the extremities of the axis are called the poles of the circle. The polas of a great circle are equally distant from the plane of the circle. The poles of a small circle are not equally distant from the plane of the circle; they may be called respectively the nearer and further pole; sometimes the nearer pole is for brevity called the pole. 6. A pole of a circle is equally distant from every point of the circumference of thie circle. Let O be the Eentre of the sphere, AB any circle of the sphere, C the centre of the circle, P and P the poles of the circle. Take any point D in the circumference of the circle; join CD, OD, PD. Then PD-= (PC+CD'); and PC and CD are constant, therefore PD is constant. Suppose a great circle to pass through the points P and D; then the chord PD is constant, and therefore the arc of ·GREAT, AND SMALL CIRCLES. 3 a great circle intercepted between P and D is constant for all positions of D on the circle AD. B С P" Thus the distance of a pole of a circle from every point of the circumference of the circle is constant, whether that distance be theast red by the straight line joining the points, or by the arc of a great circle intercepted between the points. 7. The arc of a great circle which is drawn from a pole of a great circle to any point in its circumference is a quadrant. B Let P be a pole of the great circle ABC; then the arc PA is & quadrant. For let O be the centre of the sphere, and draw PO. Then PO is at right angles to the plane ABC, because P is the pole of ABC, therefore POA is a right angle, and the arc PA is a quadrant. B 2 1 ·GREAT AND SMALL CIRCLES. 8. The angle subtended at the centre of a sphere by the arc of a great circle which joins the poles of two great circles is equal to the inclination of the planes of the great circles. B D D Let O be the centre of the sphere, CD, CE the great circles intersecting at C, A and B the poles of CD and CE respectively. Draw a great circle through A and B, meeting CD and CE at M and I respectively. Then AO is perpendicular to OC, which is a straight line in the plane OCD; and BO is perpendicular to OC, which is a straight line in the plane OCE; therefore OC is perpendicular to the plane AOB (Euclid, XI. 4); and therefore OC is perpendicular to the straight lines OM and ON, which are in the plane AOB. Hence MOV is the angle of inclination of the planes OCD and OCE. And the angle AOB = AOM - BOM = BON-BOM = MON. 9. By the angle between two great circles is meant the angle of inclination of the planes of the circles. Thus, in the figure of the preceding Article, the angle between the great circles CD and CE is the angle MON. In the figure to Art. 6, since PO is perpendicular to the place ACB, every plane which contains PO is at right angles to the plane ACB. Hence the angle between the plane of any circle And the plane of a great circle which passes through its polęs is A right angle.

·GREAT AND SMALL CIRCLES. 10. Two great circles bisect eacă other. For since the plane of each great circle passes through the centre of the sphere, the line of intersection of these planes ist diameter of the sphere, and therefore also a diameter of each great circle; therefore the great circles are bisected at the points, where they meet. 11. If the arcs of great circles joining a point P on the surface of a sphere with two other points A and C on the surface of the sphere which are not at opposite extremities of a diameter be each of them equal to a quadrant, P is a pole of the great circle through A and C. (See the figure of Art. 7.) För suppose PA and PC to be quadrants, and 0 the centre of the sphere; then since PA and PC are quadrants, the angles POC and: P04 are right angles. Hence PO is at right angles to the plane AOC, and P is a pole of the great circle AC. 12. Great circles which pass through the poles of a great circle are called secondaries to that circle. Thus, in the figure of Art. 8 the point C is a pole of ABMN, and therefore CM and CN are parts of secondaries to ABMN. And the angle between CM and CN is measured by MN; that is, the angle between any two great circles is measured by the arc they intercept on the great circle to which they are secondaries. 13. If from a point on the surface of a sphere there can be drawn two arcs of great circles, not parts of the same great circle, the planes of which are at right angles to a given circle, that point is a pole of the given circle. Fon since the planes of these arcs are at right angles to the plane of the given circle, the line in which they intersect is perpendicular to the plane of the given circle, and is therefore the Axis of the given circle; hence the point from which the arcs are drawn is a pole of the circle.

·:6 SPHERICAL TRIANGLES. 14. To compare the arc of a small circle subtending any angle at the centre of the circle with the arc of a great circle subtending the same angle at its centre. c B Let ab be the arc of a small circle, C the centre of the circle, R the pole of the circle, O the centre of the sphere. Through draw the great circles Rad and PbB, meeting the great circle of which P is a pole, at A and B respectively; draw Cd, Cb, 04, OB. Then Ca, Cb, 04, OB are all pri pendicular to OP, because the planes aCb and AOB are perpendicular to OP; therefore Ca is parallel to 0A, and Cb is parallel to OB. Therefore the angle aCb = the angle 40B (Euclid, xi. 10). Hence, 1 arc ab arc AB (Plane Trigonometry, Art. 18); radius Ca radius 0 A therefore, arc ab arc AB Ca VA re Ca Ou sin POa. II, SPHERICAL TRIANGLES. 15. Spherical Trigonometry investigates the relations which subsist between the angles of the plane faces which form a solid angle and the angles at which the plane faces are inclined to each. othor, ·SPHERICAL TRIANGLES. 7 16. Suppose that the angular point of a solid 'angle is made the centre of a sphere; then the planes which form the solid angle will cut the sphere in arcs of great circles. Thus a figure will be formed on the surface of the sphere which is called a spherical triangle if it is bounded by three arcs of great circles; this will be the case when the solid angle is formed by the meeting of three plane angles. If the solid angle be formed by the meeting of more than three plane angles, the corresponding figure on the surface of the sphere is bounded by more than tliree ares of great wrcles, and is called a spherical polygon. 17. The three arcs of great circles which form a spherical triangle are called the sides of the spherical triangle; the angles formed by the arcs at the points where they meet are called the angles of the spherical triangle. (See Art. 9.) 18. Thus, let O be the centre of a sphere, and suppose a solid angle formed at O by the meeting of three plane angles. Let , 3 AB, BC, CA be the arcs of great circles in which the planes cirt the sphere; then ABC is a spherical triangle, and the arcs AB, BC, CA are its sides. Suppose Ab the tangent at A to the arc AB, and Ac the tangent at A to the arc AC, the tangents being drawn from A towards B and C respectively; then the angle bac is one of the angles of the spherical triangle. Similarly angles formed in like manner at B and C are the other angles of the spherical triangle.

·8 SPHERICAL TRIANGLES. 19. The principal part of a treatise on spherical trigonometry consists of theorems relating to spherical triangles ; it is therefore necessary to obtain an accurate conception of 4 spherical triangle and its parts. It will be seen that what are called sides of a spherical triangle are really arce of great circles, and these arcs are proportional to the three plane angles which form the solid angle corresponding to the spherical triangle. Thus, in the figure of the preceding Article, the arc AB forms one side of the sphericr] triangle 4 BC, and the plane angle AOB is measured by the fraction radius o A ; and thus the arc AB is proportional to the angle 10B so long as we keep to the same sphere. arc AB The angles of a spherical triangle are the inclinations of the plane faces which form ide solid angle ; for since Ab and Ac are both perpendicular to 0 A, the angle bAc is the angle of inclination of the planes OAB and OAC. 20. The letters A, B, C are generally used to denote the angles of a spherical triangle, and the letters a, b, c are used to denote the sides. As in the case of plane triangles, A, B, and C may be used to denote the numerical values of the angles expressed in terms of any unit, provided we understand distinctly what the unit is. Thus, if the angle C be a right angle, we may say that C = 90°, or that C= , according as we adopt for the unit a degree or the angle sıbtended at the centre by an arc equal to the radius. So also, as the sides of a spherical triangle are oportional to the angles subtended at the centre of the sphere, we may use a, b, c to denote the numerical values of those angles in terms of any unit. We shall usually suppose both the angles and sides of a spherical triangle expressed in circular measure. (Plane Trigonometry, Art. 20.)

·SPHERICAL TRIANGLES. 9 21. In future, unless the contrgry be distinctly stated, any arc drawn on the surface of a sphere will be supposed to be an aro of a great circle. 22. In spherical triangles each side is restricted to be less than a semicircle; this is of course a convention, and it is adopted because it is found convenient, F B D Thus, in the figure, the arc ADEB is greater than a semicircumference, and we might, -if we pleased, consider ADEB, AC, and BC as forming a triangle, having its angular points at A, B, and C. But we agree to exclude such triangles from our consideration; and the triangle having its angular points at A, B, and C, will be understood to be that formed by AFB, BC, and CA. 23. From the restriction of the preceding" Article it will follow that any angle of a spherical triangle is less than two right angles. For suppose a triangle formed by BC, CA, and BEDA, having the angle BCA greater than two right angles. Then suppose D to denote the point at which the arc BC, if produced, will meet AE;hen BED is a semicircle by Art. 10, and therefore BEA is greater than a semicircle; thus the proposed triangle is not one of those which we consider, ·( 10 ) Ki II. SPHERICAL GEOMETRY. 24. The relations between the sides and angles of a Spherical Triangle, which are investigated in treatises on Spherical Trigonometry, are chiefly such as involve the Trigonometrical Functions of the sides and angles. Before proceeding to these, however, we shall collect, under the head of Spherical Geometry, some theorems which involve the sides and angles themselves, and not their trigonometrical ratios. 25. Polar triangle. Let ABC be any spherical trinngle, and let the points A', B', C' be those poles of the arcs BC, C4, 18 C B c respectively which lie on the same sides of them as the opposite angles A, B, C; then the triangle A'B'C' is said to be the polar triangle of the triangle A BC. Since there are two poles for each side of a spherical triangle, eight triangles can be formed having for their angular points poles of the sides of the given triangle; but there is only one triangle in which these poles A', B', C' lie towards the same parts with the corresponding angles A, B, C;, and this is the triangle which is known under the name of the polar triangle. The triangle ABC is called the primitive triangle with respect to the triangle A'B'C'.

·SPHERICAL GEOMETRY. 11 26. If one triangle be the polar friangle of another, the latter will be the polar triangle of the former. Let ABC be any triangle, A'B'C' the polar triangle: then ABC will be the polar triangle of A'B'C'. C D B c since For since B' is a pole of AC, the arc AB' is a quadrant, and Ć is a pole of BA, the arc AC' is a quadrant (Art. 7); there fore A is a pole of B'C' (Art. 11). Also À and A' are on the same side of B'C'; for A and A' are by hypothesis on the same side of BC, therefore A'A is less than a quadrant; and since A is a pole of B'C', and AA' is lessothan a quadrant, A and A' are on the same Bidle of B'C'. Similarly it may be shewn that B is a pole of C'A', and that B and B' are on the same side of C'A'; also that C is a pole of A'B', and that C and Care on the same side of A'B. Thus ABC is the polar triangle of A'B'C'. 27. The sides and angles of the polar triangle are respectively the supplements of the angles and sides of the primitive triangle. For let the arc B'C', produced if necessary, meet the arcs AB, AL', produced if necessary, at the points D and E respectively; then since A is a pole of B'C", the spherical angle A is measured by the arc DE (Art. 12). But B'E and C'D are each quadrants; therefore DE and B'C' are together equal to a semicircle; that is, the angle subtended by B'C' at the centre of the sphere is the ·12 SPHERICAL GEOMETRY. . supplement of the angle 4, This we may express for shortness thus; B'C" is the supplement of A. Similarly it may be shewn that C'A' is the supplement of B, and A'B' the supplenient of C. And since ABC is the polar triangle of A'B'C', it follows that BC, CA, AB are respectively the supplements of A', B', C'; that is, 4', B', C' are respectively the supplements of BC, CA, AB. From these properties a primitive triangle and its polar triangle are sometimes called supplemental triangles. Thus, if A, B, C, a, b, c denote respectively the angles and the sides of a spherical triangle, all expressed in circular measure, and A', B, C', a', 0, d' those of the polar triangle, we have A' =-a, B =-6, C'=T-C, a'=-A, =T-B, d=T-C. 28. The preceding result is of great importance; for if any general theorem be demonstrated with respect to the sides and the angles of any spherical triangle it holds of course for the polar triangle also. Thus any such theorem will remain true when the angles are changed into the supplements of the corresponding sides and the sides into the supplements of the corresponding angles. We shall see several example of this principle in the next Chapter J 29. Any two sides of a spherical triangle are together greater than the third sile. (See the figure of Art. 18.) For any two of the three plane angles which form the solid angle at O are together greater than the third (Euclid, xl. 20). Therefore any two of the arcs AB, BC, CA, are together greater than the third. From this proposition it is obvious that any side of & spherical triangle is greater than the difference of the other two. 30. The sum of the three sides of a spherical triangle is loss than the circumference of a great cirole. (See the figure of Art. 18.)

·SPHERICAL GEOMETRY. 13 + For the sum of the three plane angles which form the solid angle at O is less than four right angles (Euclid, xi. 21); therefore AB BC CA is less than 2, OAOA OA therefore, AB + BC + C'A is less than 24 * OA; that is, the sum of the arcs is less than the circumference of a great circle, 31. The propositions contained in the preceding two Articles may be extended. Thus, if there be any polygon which has each of its angles less than two right angles, any one side is less than the sum of all the others. This may be proved by repeated use of Art. 24 Suppose, for example, that the figure has four sides, and "let the angular points be denoted by A, B, C, D. Then AB + BC is greater than AC; therefore, AB + BC + CD is greater than AC + CD, and à fortiori greater than 4D. Again, if there be any polygon which has each of its angles less than two right angles, the sun of its sides will be less than the circunference of a great circle. This follows from Euclid, XI. 21, in the manner shewn in Art. 30, 32. The three angles of a spherical triangle are together greater than two right angles and less than six right angles. Let A, B, C be the angles of a spherical triangle; let a', 6, .' be the sides of the polar triangle. Then by Art. 30, á' + b + d is less than 27, that is, H-A+#-B +-C is less than 2w; therefore, A+B+C is greater than . And since each of the angles A, B, C is less than 7, the sum A+B+C is less than 37. .

·14 SPHERICAL GEOMETRY. 33. The angles at the bage of an isosceles spherical triangle are equal. f B T Let ABC be a spherical triangle having AC = BC; let O be the centre of the sphere. Draw tangents at the points A and B to the arcs AC and BC respectively; these will meet OC produced at the same point S, and AS will be equal to BS. Draw tangents AT, BT at the points A, B to the arc B; then AT = T'B; join T's. In the two triangles SAT, SBT the sides SA, AT, TS are equal to SB, BT, T's respectively; therefore the angle SAT is equal to the angle SBT ; and these are the angles at the base of the spherical triangle. The figure supposes AC and BC to be less than quadrants; if they are greater than quadrants the tangents to AC and BC will meet on CO produced through O instead of through C, and the demonstration may be completed as before. If AC and BC are quadrants, the angles at the base are right angles by Arts. 11 and 9. 02 34. If two anglès of a spherical triangle are equal, the opposits sides are equal. Since the primitive triangle has two equal angles, the polar triangle has two equal sides; therefore in the polar triangle the angles opposite the equal sides are equal by Art. 33. Hence in the primitive triangle the sides opposite the equal angles , are equale ·SPHERICAL GEOMETRY. 15 35. If one angle of a spherical triangle be greater than another, the side opposite the greater angle is greater than the side opposite the less angle. Let ABC De a spherical triangle, and let the angle ABC be Greater than the angle BAC : then the side AC will be greater than the side BC. At B make the angle ABD equal to the angle DAD; then BD is equal to AD (Art. 34), and BD+DC is greater tu BC (Art. 29); therefore AD.+ DC is greater than BC; that, is, AC is greater than BC. 36. If one side of a spherical triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less side. This follows from the preceding Article by means of the polar triangle. Or thus; suppose the side AC greater than the side BC, then the angle ABC will be greater than the angle BAC. For the angle ABC cannot be less than the 'angle BAC by Art. 35, and the angle ABC cannot be equal to the angle BAC by Art, 34; therefore the angle ABC must be greater than the angle BAC. This Chapter might be extended; but it is unnecessary to do so because the Trigonometrical formulæ of the next Chapter supply an easy method of investigating the theorems of Spherical Geometry. See Arts, 56, 57, and 58.

·( 16 )
IV. RELATIONS BETWEEN THE TRIGONOMETRICAL
FUNCTIONS OF THE SIDES AND THE ANGLES,
OF A SPHERICAL TRIANGLE.
37. To express the cosine of an angle of a triangle in terms of
sines and cosines of the sides.
D
B
Let ABC be a spherical triangle, O the centre of the sphere.
Let the tangent at A to the arc AC' meet OC produced at E, anul
let the tangent at A to the arc AB meet OB produced at D; join
ED. Thus the angle EAD is the angle A of the spherical triangle,
and the angle EOD measures the side a.
From the triangles ADE and ODE we have
DE’= ADS + AE - 2AD. AE cos A,
DE = OD' + OE – 200.0E cosa;
also the angles OAD and CAE are right angles, so that
OD=0A + ADS and OE' =0A + AE. Hence by subtraction
we have
0= 204+ 2AD. AE cos A-20D.OE cosa;
OA OA AE AD
therefore
OE OD JE OD
cos A;
that is cos a = cos 6 cos c + siņ 6 sin c cos A.
cos Q - COB 6 cos C
Therefore
sin b sino
cOS a=
COS A =

·RELATIONS BETWEEN THE FUNCTIONS. 17 38. We have supposed, in the construction of the preceding Article, that the sides which contain the angle A are less than quadrants, for we have assumed that the tangenta at A meet OB apd OC respectively produced. We must now shew that the formula obtained, is true when these sides are not less than quadrants. This we shall do by special examination of the cases in which one side or each side is greater than a quadrant or equal to u quadrant. (1) Suppose only one of the sides greater than a quadrant, for example, AB. Produce BA and BC to meet at B'; and put 'C', CB=d'. B. B Then we have from the triangle AB'C, by what has been ABC alrc.uly provel, cos a' = cos 6 cos c' + sin b sin c'cos B'AC; but a a, c' = 9 - C, D'AC=T thus cos a = cos 6 cos C + sin b sin c cos An . -A; (2) Suppose both the sides which contain the angle 1 to be greater than quaurants. Produce AB and AC to meet at A'; put I'B-6, A'C=b'; then from the triangle A' BC, as before, В. cos a = cos b' cos d' + sin b' sin c' cos c'cos A'; but (= 1 - 6, c'=7-C, A=A ; thus cos Q = cos 6 cos o + sin b sin c cos A. T. S. T.

·18 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS (3) Suppose that one of the sides which contain the angle A is a quadrant, for example, AB; on AC, produced if necessary, AA 7 as well TT as C . 7 cos CD COS 6 COS A take AD equal to a quadrant and draw BD.. If DD is a qualront B is a pole of AC (Art. 11); in this case a -- and A 2 2 Thus the formula to be verified reduces to the identity 2 0= 0. If BD be not a quadrant, the triangle BDC gives cos a = cos CD cos BD + sin CD sin BD cos CDB, and cos CDB = 0, )sin b, cos BD = cos il; thus sin b coś A ; and this is what the formula in Art. 37 becomes when c= 2 (4) Suppose that both the sides which contain the angle A are quadrants. The formula then becomes cos a = cos A; and this is obviously true, for A is now the pole of BC, and thus A Thus the formula in Art. 37 is proved to be universally true. 39. The formula in Art. 37" may be applied to express the cosine of any angle of a triangle in terms of sines and cosines of the sides; thus we have the three formulæ, cos a = cos 6 cos c + sin b sin c cos A, = cos c cos a + sin c sin a cos B, cos C= cos a cos 6 + sin a sin b cos C. =a. cos 6 ·OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 19 These may be considered as the funda hental equations of Spherical Trigonometry; we shall proceed to deduce various formula from them. 40. To express ths sine of an angle of a spherical triangle in terms of trigonometrical functions of the sides. cos i cos c We havo sin 6 sin c COS ( COS A ; 2 therefore sin" A=1 COS a cos 6 cos c sin o sin c 99 cos 6 cos c)? (1 - cos®6) (1 - cos c) – (cos a siu*. sinc 1-cos'a - cos'b - cosc + 2 cos a cos b cos c sin'b sinc (1-cos*a-cos’l-cosºc+2 cos a cos 6 cos e) therefore sin A sin b sin c The radical on the right-hand side must be taken with the positive sign, because sin ), sin c, and sin A are all positive. 41. From the value of sin A in the preceding Article it fullows that sin sin B sin C sin c Sil u sin , for each of these is equal to the same expression, namely, 1 (1 - cos'a - case6 - cos'c + 2 cos acosb cosc) sin a sin b sinc Thus the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. We will give an independent proof of this proposition in the following Article. 49. The sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. C2 ·20 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS Let ABC'be a spherical triangle, 0 the centre of the sphere. Take any point P in 01, draw PD perpendicular to the plane В. D F BOC, and from D draw DE, DF perpendicular to OB, OC respectively; join PE, PF, OD. Since PD is perpendicular to the plane BOC, it makes right angles with every straight line moeting it in that plane; horico PE`=PDS + DE = POP – OD + DES = POP - OF?; thus PEO is a right angle. Therefore PE-OP sin POE=0P sinc; and PD=PE sin PED=PE sin B=OP sin c sin B. Similarly, PD = OP sin b sin C; therefore OP sin c sin B=0 sin / sin C; sin B sin b therefore sin C sin c' The figure supposes b, c, B, and C each less than a right angle; it will be found on examination that the proof will hold when the figure is modified to meet any case, which can If, for instance, B alone is greater than a right angle, the point D will fill beyond OB instead of between OB and OC ; then PED will Le the supplement of B, and thus sin PED is still equal to sin B. 43. To shew that cot a sin b = cot A sin C + cos 6 cos C. We have cos a=cos 6 cos C+ sin b sin coos, COS C=cos a cos 6 + sin a sin b cos C, sin C sin c = sin a sin A occur.

·OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 21 cos a = 2 Substitute the values of cosc and sinc in the first equation ; thus sin a sin 6 cos A sin C (cos a cos 6 + sin a sin b cos C) cos b + sin A by transposition cos a sin*b = sin a sin b cos b cos (+ sin a sin b cot A sin C;, divide by sin a sin b; thus cot a sin b = cos 6 cos C + cot A sin C: '44. By interchanging the letters five other formulæ may be obtained like that in the preceding Article; the whole six formule will be as follows: cota sin b = cot A sin C + cos b cos C, cot b sin a = cot B sin C + cos a cos C, cot b sin c = cot B sin A it cos c cos A, cotc sino cot Csin A + cos b cos A, cot c sin a = cot C sin B + cos a cos B, cota sin C -- cot A sin B + cos c cos B. 45. To express the sine, cosine, and tangent, of half an angle of a triangle as functions of the sides. 3 We have, by Art. 37, COS IA cos a cos 6 cos C sin b sinc therefore 1-cos A=1. COS a cos 6 cos c sin o sin c cos (6-c) - cosa sin 6-sinc i A sin } (a+b-c) sin } (a −6+c). therefore sinº sin 6 sin c Let 28 = a + b + c, so that 8 is half the sum of the sides of the triangle; then a +6-6= 28 - 2c=2(8-c), a-6+c=28 - 26 = 2(8-5); thus sin 4 sin (8— b) sin (8- c) 2 sin 6 sin c ·22 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS and sin (sin (8-6) sin sin l siuc 3 sin –0). 1 COS - Also, 1 + cos A=1+ COS ( cos 6 cos c sin b sin c cos (b + c) sin b sin c therefore A cos 3 9 sin } (a + + + c) sin } (6+c-a)_ sin 8 sin (8 — «) sind sinc sin sinc > A and COS Vision sin 8 sin (8 - 02 siu o sin c From the expressions for sin А 2 and cos 2 we deduce tan A 2 (sin (s – b) sin (s -c)) siues sin (s-u) si The positive sign nuust be given to the radicals which occur in A this. Article, because is less than a right angle, and therefore its 2 sine, cosine, and tangent are all positive. 46. Since sin A = 2 sin COS we obtain 2 2 2 sin A sin 6-sin c {sins sin (8-– a) sin (8 --U) sin (8–c)}4. It may be shewn that the expression for sin A in Art. 40 agrees with the present expression by putting the numerator of Ithat expression in factors, as in Plane Trigonometry, Art. 115. We shall find it convenient to use a symbol for the radical in the value of sin A : we shall denote it by n, so that no = sin 8 sin (8-a) sin (8-6) sin (8-c), and An=1 - cosa - cos'b-cos'c + 2 cos a cos 6 cos C.

·OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 23 47. To express the cosine of a side of a triangle in terms of sines and cosines of the angles. In the formula of Art. 37 we may, by Art. 28, change thell sides into the supplements of the corresponding angles and the ingle into the supplement of the corresponding side; this Cos (7 - 1) = cos (71 - BYcos (T-C) + sin (-B) sin (7 - C') cos(-a), that is, cos B cos C + sin B sin C cos a.) Similarly cos C cos A + sin C sin A cos b, ind cos A cos B + sin A sin B cos c. cOS A COS B cos C E w 48. The forinulæ in Art. 44 will of course remain true when ingles anil sides are changed into the supplements of the corl'esponding sites and angles respectively; it will be found, however, that no new forinula are thus obtaineil, but only the same formula over again. This consideration will furnish some assists ance in rytaining those formulæ accurately in the memory. 49. To express the sine, cosine, and tangent, of half a side of a triangle as functions of the unyles. cos 1 + cos B cos C We have, by Art. 47, sin B sin COS a therefore 1- cos a = 1 cos A + cos B cos C sin B sin c cos A + cos (B+C) sin B sin ; therefore sinº cos } (A + B+C) cos (B+C - A) sin B sin c Lot 28 = A + B +C; then B+C-A=2(8-4), therefore a sin? cos .S cos (S - A) siu B sin c > 2 COS and sin = { s cos (S - A) sin B sin c 13 ·24 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS cos cos + cos B cos C cos A + cos (B-C) Also 1 + COS Q = 1+ sin B sin sin B sin c therefore cos (A-B+C) cos (A + B -~_cos (S – B) cos (S-C) sin B sinc sin B sinc cos (S - B) cos (S-C)) and 2 sin B siuc cos S cos (S - A) Hence tan cos (8 – B) cos (8-C) The positive sign must be given to the radicals which occur in this Article, because is less than a right angle. a COS Dicas VA 2011! (L 50. The expressions in the preceding Article may also be. obtained immediately from those given in Art. 45 by means of Art. 28. a (1 It may be remarked that the values of sin COS and tan 2 T are real. For S is greater than one right angle and less than three right angles by Art, 32; therefore cos S is negative. And in the polar triangle any side is less than the sum of the other two; thus 1 - A is less than 5 B+-C; therefore B+C - A is less than 7; therefore S-A is less than and B+C-A is algebraically 2' greater than - , so that S-A is algebraically greater than 2; therefore cos ( SA) is positive. Similarly also cos (S-B) and cos(S-C) are positive. Hence the values of sin 2 sa, and tan a COS are real. sin a 61. Since sin a = 2 sin cosa, we obtain cos S cos (S - A) cos (S - B) cos (S --014 We shall use N for {-cos S cOS (S- A) COB (S - B) cos (S-2)}t. 2 sin B sin c ·OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 23 52. To demonstrate Napier': Anwlogies. sin A We have sin B sin 6 = m suppose ; sin a then, by a theorem of Algebra, m sin A + sin B sin u + sin 6 -$1), ...... sin A-sin B and also m = .(2). sill a 8116 Now cos A + cos B cos C = sin B sin C cosa= m sin C sin b cosa, and cos B + cos A cos C = sin A sin C cos b = m sin C sin a cos b, therefore, by addition, (cos A + cos B) (1 + cos C) = m sin sin (a + b) ..........(3); therefore by (1) we have sin A + sin B sin a + sin l 1 + cos C COS & + coy B siu ( + 6) Sul cos } (a - b) that is, tan 2(A + B) cost (a + ) cot 9 3 sin a Similarly from (3) and (2) we have sin A -sin B - sin b *1 + cos C COS A + cos B sin (a + b) sin sin (a-6) tan (A-B) cot sin(a+b) o that is, 2013 (5). By writing - A for a, and so on in (4) and (5) we obtain cos } (4 - B tan tan (a + b)= ..(6), cos (4+B) sin }(A - B) tan (a - b) = tan sin (A + B) Lolo Lors .....(7) The formule (4), (5), (6), (7) may be put in the form of proportions or analogies, and are called from their discoverer Napier's ·< 26 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS Analogies: the last two may be demonstrated without regarring to the polar triangle by starting with the formulæ in Art. 39. 53. In equation (1) of the preceding Article, cos } (a - b) une 0 cot 2 are necessarily positive quantities; hence the equation shows that tan | (A + B) and cos } (a + b) ere of the same sign ; thus 3 (A + B) and (a + b) are either both less than a right angle or both greater than a right angle. This is expressed by saying that 14+B) and (a + b) are of the same affection. Ci + 54. To demonstrate Delambre's Analoyies. We have cos c= cos a cos b + sin a sin l cos C; therefore, 1 + cos C= 1 + cos a cos 6 + sin a sin b (cos' C - sinº 1 C). ={1+cos (a−b)}cos* } C +{1+cos (a+)} sin?, therefore cos’ c= cos' } (a - ) cos' { C + cos® ! (a +h) sin' } /. Similarly, sino { c == sinº ) (a-b) cos' 1 C+sin' ! (a + b) sin" 16. Now add unity to the square of each member of Napier's first two analogies; hence by the formulæ just proved sec" } (A+B) - cos" ! (a + b)sin}C? cos' dc sinc sec" } (A - B) = sin" (a + O) sin" tc · Extract the square roots ; thus, since (A + B) and 1 (a +6) are of the same affection, we obtain cos (A + B) cos c= cos } (a + b) sin C ......... (1), cos 5 (4 – B) sin fc sin : (a + b) sin of C ......... (2). Multiply the first two of Napier's analogies respectively by these results; thus sin $ (A + B) cos fc = cos } (a - b) cos C......... (3), sin (4 – B) sin $c= sin d (a - b) cos C......... (t).

·OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 27 The last four formule are commonly, but very improperly, called Gauss's Theorems; they were first given by Delambre in the Connaissance des Tems for 1809, page 445. 55. The properties of supplemental triangles were proved geometrically in Art. 27, and by means of these properties the formulæ in Art. 47 were obtained ; but these formulæ may be leduced analytically from those in Art. 39, and thus the whole subject may be mue to depend on the formulæ of Art. 39. For from Art. 39 we obtain expressions for cos A, cos B, cos C; and from thicse we find cos A + Ços B cos C (cos a - cos b cos c) sin' a + (cos b- cos a cos c)(cos c cos a cos 7) sinº a siu b sin c In the numerator of this fraction write 1 - cos'a for sina; this the immerator will be found to reduce to ÇOS (4 (1 - cos*a -. cos 6 – cos'c + 3 cos a cos 6 cos c), ind this is equal to cos a sin B sin C'sin' a sin b sin c, (Art. 41); therefore COS A + cos B cos C = cos a sin B sin C. Similarly the ouer two corresponding formulæ may be proved. Tiuus the formulæ in Art. 47 are established; and therefore, without assuming the existenre and properties of the Polar Triangle, we deduce the following theorem: If the sides and angles of a spherical triangle be changed respectively into the supplements of the corresponding angles and sides, the fundamental formula of Art. 39 hold good, and therefore also all results deducible from them. 56. The formulæ in the present Chapter may be applied to establish analytically varions propositions respecting spherical triangles which either have been proved geometrically in the preceding Chapter, or may be so proved. Thus, for example, the Lecond of Napier's analogies is sin (a-6) tan (A - B) cot sin $ (a + b) ge ;

·28 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS this shews that }(4-B) is positive, negative, or zero, according as (a - b) is positive, negative, or zero; thus we obtain all the results included in Arts. 33...36. 57. If two triangles have two sides of the ope equal to two sides of the other, each to each, and likewise the included angles equal, then their other angles will be equal;each to each, and likenoise their bases will be equal. We may she that the bases are equal by applying the first formula in Art. 39 to each triangle, supposing b, c, and a tha same in the two triangles ; then the remaining two formulæ of Art. 39 will shew that B and C are the same in the two triangles. It should be observed that the two triangles in this case are not necessarily such that one may be made to coincide with the other by superposition. The sides of one may be equal to those of the other, each to each, but in a reverse order, is in the follywing figures. ☺ Two triangles which are equal in this manner are said to be symmetrically equal; when they are equal so as to admit of superposition they are said to be absolutely equal. * * 58. If two spherical triangles have troo sides of the one equal to two sides of the other, each to each, but the angle which is contained by the troo sides of the one greater than the angle which is contained by the two sides which are equal to them of the other, the base of that ·OF THE SIDES AND THE RNGLES OF A SPHERICAL TRIANGLE. 29 which has the greater angle will beagreater than the base of the other, and conversely. Let 6 and c deilote the sides which are equal in the two triangles; let a be the base and A the opposite angle of one triangle, and a' and A' similar quantities for the other. Then cos a --- cos 6 cos C + sin b sin c cos A, = cos 6 cos c+sin b silec cos therefore cos a' = sin b sin c (cos A - cos'A'); that is, sin } (a + a') sind (a-a')= sin b sin c sin } (A + A') sin (A - '); this shews that } (a --a') and 1 (4 - A) are of the same sign. cos a' COS a 59. If on a sphere any point be taken within a circle which! is not its pole, of all the arcs whith can be drawn from that point to the circunference, the greatest is that in which the pole is, and the other part of that produced is the least ; and of any others, that which is nearer to the greatest is always greater than one more remote; and from the same point to the circumference there can be drawn only two arcs which are equal to esch other, and these make equal angles with the shortest arc on opposite sides of it. This follows readily from the preceding three Articles, 60. We will give another proof of the fundamental formula in Art. 39, which is very simple, requiring only a knowledge of the elements of Co-ordinate Geometry. Suppose ABC any spherical triangle, O the centre of th sphere, take 0 as the origin of co-ordinates, and let the axis of 2 pass through C. Let 2, 4, z, be the co-ordinates of A, and x, 92, %, those of B; let r be the radius of the sphere. Then the square on the straight line AB is equal to (, -2) + (y - y)+(2,-2)', and also to gol + goed - 2rcos A0B;

·30 EXAMPLES. and *,*+7,42 * = *, * ny * +2 = , thus 2,3, +7,7, +26,= gol cos AOB. Now make the usual substitutions in passing from rectangular to polar co-ordinates, namely, %, = r cos 0,, , 2,=r sin 8, cos , y=r sin 6, sin ,, %, =r cos 2nr sin 0,cos de, y,=r sin ,sin ,; thus we obtain C cos , cos 0, + sin 0, sin , cos ($.-.) = COS AOB, that is, in the ordinary notation of Spherical Trigonometry, cos a cos b + sin a sin b cos C This method has the advantage of giving a perfectly general proof, as all the equations used are universally true. = COS C. EXAMPLES. Jn. If A = a, shew that B and b are equal or supplemental, as also C and c. 2. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle. 3. When does the polar triangle coincide with the primitive triangle? 4. If D be the middle point of AB, shew that cos :4C + cos BC = 2 cos } AB cos CD. 5. If two angles of a spherical triangle be respectively equal to the sides opposite to them, shew that the remaining side is the supplement of the remaining angle; or else that the triangle has two quadrants and two right angles, and then the remaining side is equal to the remaining angle.

·• EXAMPLES. 31 6. In an equilateral triangle, shew that 2 cos sin A 2 1. a 7. In an equilateral triangle, shew that tan 2 1-2 cos A ; -1- hence deduce tlte limits between which the sides and the angles of all equilateral triangle are restricted. 8. In an equilateral triangle, shew thidt sec A = 1+ sec a. C 29. If the three sides of a spherical triangle be halved and 6 a new triangle formed, the angle 0 between the new sides and 2 6 is given by cos 0 = cos A + įtan tan sinº 0. C 10. 1B, CD are qualrants on the surface of a sphere intersecting at E, the extremities being joined by great circles : shew. that cos 1 EC = cos AC cos BD - cos BC cos AD. 11. If 6+ c = show that sin 2B + sin 2C =0. 12. If DE be an arc of a great circle bisecting the sides AB, AC of a spherical triangle at D and E, Pa pole of DE, and PL, PD, PE, PC be joined by arcs of great circles, shew that the angle BPC=twice the angle DPE. 13. Shew that B cos C cos a. sin b sin c.+ cos cos C COS A = sin B sin C - cos 14. If D be any point in the side BC of a triangle, shew that cos AD sin BC = COS AB sin DC + cos AC sin BD. 915. Shew that if 0, 0, y be the lengths of arcs of great circles drawn from A, B, C perpendicular to the opposite sides, sin a sin 0 = sin o sin $ = sin c sin y - 1 (1 - cos' a - cos' 6 - cos* c + 2 cos a cos 6 cos c).

·32 SOLUTION OF RIGHT-ANGLED TRIANGLES. 16. In & spherical triangle, if 0, 0, 4 be the arcs bisecting the angles A, B, C respectively and terminated by the opposite sides, shew that A B с cot A cos + cot $ cos + coty cos cot a + cot b + cot c. 2 37. Tavo ports are in the same parallel of latitude, thoir common latitude being 1 and their difference of longitude 2: shew that the saving of distance in sailing from one to the other on the great circle, instead of sailing due East or West, is 2r { cos 1-sin-' (sin cos)}, 2 being expressed in circular measure, and r being the rulius of the Earth 18. If a ship be proceeding uniformly along a great circle and the observed latitudes be li, 1,2 lg, at equal intervals of time, in each of which the distance traversod is 8, slew that sin } (l. +?) cos d (1.-?) 8 = r cos-1 siul, r denoting the Earth's radius: and shew thing the change of longi. tude may also be found in terms of the three latitudes. V. SOLUTION OF RIGHT-ANGLED TRIANGLES. 61. In every spberical triangle there are six elements, namely, the three sides and the three angles, besides the radius of the sphere, which is supposed constant. The solution of spherical triangles is the process by which, when the values of a sufficient number of the six elements are given, we calculate the values of the remaining elements. It will appear, as we proceed, that when the values of three of the elements are given, those of the remain- Jing three can generally be found. We begin with the right-angled triangle: here two elements, in addition to the right angle, will be supposed known.

·SOLUTION OF RIGHT-ANGLED TRIANGLES, 33 62. The formulæ requisite for the solution of eight-angled triangles may be obtained from the preceding Chapter by supposing one of the angles a right angle, as C for example. They måy also be obtained very easily in an independent manner, as. we will now shew. N B Let ABC be a spherical triangle having a right angle at C;. let () be the centre of the sphere. From any point P in OA draw PM perpendicular to OC, and from M draw MN perpendicular to OB, and join PN. Then PMI is perpendicular to MN, because tho plane AOC is perpendicular to the plane BOC; hence PN-P1+ UN:0P-OM?+OM:-ON=0P-On therefore P.VO is a right angle. And ON ON ON OP OM OP that is, cos c=cos a cos b............(1), PM PI PN that is, sin b = sin B sin UP POP Similarly sin A sine) (2), sin a= MY UN MN PN that is, tan a=cos B tanc PN'ON Similarly tan b = cos A tan nos (3), 3 PM PM MN OM MNOM that is, tan b= tan B sina Similarly tan a = tan A sin b T. S.T. D ·34 SOLUTION OF RIGHT-ANGLED TRIANGLES. Multiply together the tyo formulæ (4); thus, 1 tan A tan B tan a tan 6 sin a sin b 1 cos a cos 6 by (1); VOS C therefore COS C =cot A cot B..... ..(5). •••••• Multiply crosswise the second formula in (3) and the first in (3); thus sin a cosB tan c= tan a sin A sin c; sin A cos C therefore cOS BE sin A cos b by (1). COS at thus Similarly cos B =sin A cos b) cos A = sin B cosas (6). These six formulæ comprise ten equations; and thus we can solve every case of right-angled triangles. For every one of these ten equations is a distinct combination involving three out of the five quantities a, b, c, X, B; and out of five quantities only t.:n combinations of three can be formed. Thus any two of the five quantities being given and a third required, some one of the preceding ten equations will serve to determine that third quantity. 63. As we have stated, the above six formulæ may be, obtained from those given in the preceding Chapter by supposing Ca right angle. Thus (1) follows from Art. 39, (2) from Art. 41, (3) from the fourth and fifth equations of Art. 44, (4) from the first and second equations of Art. (44), (5) from the third equation of Art. 47, (®) from the first and second equations of Art. 47. Since the six formule may be obtained from those given in the preceding Chapter which have been proved to be universally true, we do not stop to shew that the demonstration of Art. 62 may be applied to every case which can occur; the student may for exercise investigate the modifications which will be necessary when we suppose one or more of the quantities a, b, c, A, B equal to a right angle or greater than a right angle.

·SOLUTION OF RIGHT-ANGLED TRIANGLES. 35 64. Certain properties of right-angled triangles are deducible from the forinulæ of Art. 62. From (1) it follows that cosc has the same sign as the product cos a cos b; henco either all the cosines are positive, or else unly one is positive. Therefore in a right-angled triangle either all the three sides are less than quadrants, or else one side is less than'a quadrant and the other two sides are greater than quadrants. Fron (1) it follows that tan a has the same sign as tan A. Thereforo A and a are either both greater than or both less T 2' T than 2 ; this is expressed by saying that A and a are of the same cffection.. Similarly B and I are of the same affection, 65. The formula of Art. 62 are comprised in the following enunciations, which the student will find it useful to remember; the results de distinguished by the same numbers as have been already applied to them in Art. 62; the side opposite the right angle is called the hypotenuse : Cos hyp = product of cosines of sides (1), Cos hyp = product of cotangents of angles ..... .(5), Sine side=sine of opposite angle x sine hyp.... ...(2), Tan side = tan hyp * cos included angle .....(3), Tan side = tan opposite angle x sine of other side.....(4), Cos angle=cos opposite side x sine of other angle...... (6). 66. Napier's Rules. The formulæ of Art. 62' are comprised in two rules, which are called, from their inventor, Napier's Rules of Circular 'Parts. Napier was also the inventor of Logarithms, and the Rules of Circular Parts were first published by him in a work entitled Mirifici Logarithmorum Canonis Descriptio...... Edinburgh, 1614. These rules we will now explain. D2 ·36 SOLUTION OF RIGHT-ANGLED TRIANGLES. The right angle is lef, out of consideration; the two sides which include the right angle, the complement of the hypotenuso, and the complements of the other angles are called the circular n C 3 T IT a, b, T parts of the triangle. Thus there are five circular parts, namely, A, - B; and these are supposed to bus ranged 2 2 2 round a circle in the order in which they naturally occur with respect to the triangle. Any one of the five parts may be selected and called middle part, then the two parts next to it are called adjacent parts, and the remaining two parts are called opposite parts. For example, if -B is selected as the middle part, then the adjacent 2 parts are a and -c, and the opposite parts are 6 and 7-4. Then Napier's Rules are the following: sine of the middle part = product of tangents of adjacent parta, Usine of the middle part=product of cosines of opposite parts, 67. Napiet's Rules may be demonstrated by shewing that they agree with the results already established the folloying table shows the required agreement: in the first column are given the middle parts, in the second column the results of Napier's Rules, and in the third column the same results expressed as in Art. 62, with the number for referend used in that article.

·SOLUTION OF RIGHT-ANGLED TRIANGLES. 37 7 cos c = cot A cot B..(5), sin = cos a cos 6 cos c = cos a cos b..(1) cos B=tan a cot c.. (5), -e min ( - ) - tan (65 – 4) tun (3-0) (4-0) 3 - B sin(5 - B)= tan a tan (1-0) sin ( - 1) = cos cos (5-4) (1 - 1) (7. – 4) cos (7-4) cos B = Cos b sin A..(6), ( sin a - tan b tan sin a=tan b cot B..(4), sina COS sin a= sin A sin c..(2), TT 6 sin b=tan A tan'a sin b = cot A tan a..(4), 3 sin b = COS sin b = sin B sin c.. (2), (5-1) cos (3 -c) -1 sin(-4)-tanótan- (-4) (-1) cos A = tan b cot c.. (3), 2 . sin -- COS CU COS COS A cos a sin B..(6). The last four cases need not have been given, since it is obvious that they are only repetitions of what had previously been given; the seventh and eighth are repetitions of the fifth and sixth, and the ninth and tenth are repetitions of the third and fourth. 68. It has been sometimes stated that the method of the preading Article is the only one by which Napier's Rules can be demonstrated; this statement, however, is inaccurate, since besides this method Napier himself indicated another method of proof in his Mirifici Logarithmorum Canonis Descriptio, pp. 32, 35. This we will now briefly explain, ·38 SOLUTION OF RIGHT-ANGLED TRIANGLES. H >D L A Let ABC be a spherical triangle right-angled nt C'; with B as pole describe a great circle DEFG, and with A as pole describe a great circle HFKL, and produce the sides of the original triangle ABC to meet these great circles. Then since B is a pole of DEFG the angles at D and G are right angles, and since A is a pole of IFKL the angles at II and L are right angles. Hence the five triangles BAC, AED, EFII, FKG, KBL are all right-angled; and moreover it will be found on examination that, although the elements of these triangles are different, yet their circular parts are the same. We will consider, for example, the triangle AED; the angle EAD is equal to the angle BAC; the side AD is the complement of AB; as the 'angles at C and G are right angles E is a pole of GC (Art. 13), therefore EA is the complement of as B is a pole of DE the angle BED is a right angle, therefore the angle AED is the complement of the angle BEC, that is, the angle A ED is the complement of the side BC (Art. 12); and similarly the side DE is equal to the angle DBE, and is therefore the complement of the angle ABC. Hence, if we denote the elements of the triangle ABC as usual by a, b, c, A, B, we have in the triangle AED the hypotenuse equal to -b, the angles equal to 2 and 2 a, and the sides respectively opposite these angles equal * B and 5-6. The circular parts of AED are therefore the ·SOLUTION OF RIGHT-ANGLED TRIANGLES. 39 same as those of ABC. Similarly the remaining three of the five right-angled triangles may be shewn to have the same circular parts as the triangle ABC has. Now take two of the theorems in Art. 65, for example (1) and (3); then the truth of the ten cases comprised in Napier's Rules will be found to follow from applying the two theorems in succession to the five triangles formed in the preceding figure. Thus this method of considering Napier's Rules regards each Rule, not as the statement of dissimilar properties of one triangle, but as the statement of similar properties of five allied triangles. 69. In Napier's work a figure is given of which that in the preceding Article is a copy, except that different letters are used; Napier •briefly intimates that the truth of the Rules can be easily seen by means of this figure, as well as by the method of induction from consideration of all the cases which can occur. The late T. S. Davies, in his edition of Dr Hutton's Course of Mathematics, drew attention to Napier's own views and expanded the demonstration by a systematic examination of the figure of the preceding Article. It is however easy to eyade the necessity of examining the whole figure; all that is wanted is to observe the connexion between the triangle A ED and the triangle BAC. For let a, a, az, a, a, represent the elements of the triangle BAC taken in order, beginning with the hypotenuse and omitting the right angle; then the elements of the triangle AED taken in order, beginning with the hypotenuse and omitting the right angle, are ag If, therefore, to characterise 2 - ,, 2 the former we introduce a new set of quantities Pu PoPsr Pu, Po, such that a, + P, = 0, +Ps = Nig+ po 2' then the original triangle being characterised by Pi, Pg, Ps, Po, po, the second triangle will be similarly characterised by , Pas por Pr, P. As the second triangle can give rise to a third in like manner, and so on, we see that every right-angled triangle is one 7T 7T IT and - dg 2 and that Pa = dg and pas ·40 SOLUTION OF RIGHT-ANGLED TRIANGLES. of a system of five such triangles which are all characterised by the quantities P P P P Po, always taken in order, each quantity in its turn standing first. The late R. L. Ellis pointed out this connexion between the five triangles, and thus gave the true significance of Napier's Rules. The memoir containing Mr Ellista investigations, which was unpublished when the first edition of the present work ap peared, will be found in pages 328...335 of The Mathematical and other writings of Robert Leslie Ellis.... Cambridge, 1863. Napier's own method of considering his Rules was neglected by writers on the subject until the late T. S. Davies drew attention to it. Hence, as we have already remarked in Art. 68, an erroneouis statement was made respecting the Rules. For instance, Woodhouse says, in his Trigonometry: “There is no separate and independent proof of these rules ;..." Airy sitys, in the treatise on Trigonometry in the Encyclopædia Metropolitana: “These rules are proved to be true only, by showing that they comprehend all the equations which we have just found." 70. Opinions have differed with respect to the utility of Napier's Rules in practice. Thus Woodhouse says, “In the whole compass of mathematical science there cannot be found, perhape, rules which more completely attain that which is the proper object of rules, namely, facility and brevity of computation." (Trigonometry, chap. x.) On the other hand may be set the following sentence from Airy's Trigonometry (Encyclopædia Metropolitana): “In the opinion of Delambre (and no one was better qualified by experience to give an opinion) these theorems are best recollected by the practical calculator in their unconnected form." See Delambre's Astronomie, vol. I. p. 205. Professor De Morgan strongly objects to Napier's Rules, and says (Spherical T'rigorometry, Art. 17): “ There are certain mnemonical formule called Napier': Rules of Circular Parts, which are generally explained. We do not give them, because we are convinced that they anly create confusion instead of assisting the memory."

·SOLUTION OF RIGHT-ANGLED TRIANGLES. 41 71. We shall now proceed to apply the formule of Art. 62 to the solution of rightangled triangles. We shall assume that the given quantities are subject to the limitations which are stated in Arts. 22 and 23, that is, a given side must be less than the semicircumference of a great circle, and a given angle less than two right angles. There will be six cases to consider. 72. Ilaving given the hypotenuse c and an angle A. Here we have from (3), (5) and (2) of Art. 62,. tan b tan c cos A, cot B = cos c tan A, sin a = sin c sin A. Thus 1 and D are determined immediately without ambiguity; and as a must be of the same affection as A (Art. 64), a also is determined without ambiguity. It is obvious from the formulæ of solution, that in this case the triangle is always possible. If . and A are both right angles, g is a right angle, and b and. Bare indeterminate. 73. Ilaring given a side b and the adjacent angle A. Here we have from (3), (4) and (6) of Art. 62, tan b tan A sin b, = cos b sin A. Here c, a, B are determined without ambiguity, and the triangle is always possible. tan c= tan a= cos B cos A' COS C 74. Having given the two sides a ani b. Here we have from (1) and (4) of Art. 62, cos a cos b, cot A=cot a sin b, cot B=cot b sin a. Here C, A, B are determined without ambiguity, and the triangle is always possible. 75. 'Ilaving given the hypotenuse c and a side a Here we have from (1), (3) and (2) of Art. 68, tan a COS C sina cos 6 cos B= sin A: > > CO8 a tan c Sin c ·SOLUTION OF RIGHT-ANGLED TRIANGLES. Here , B, A are determined without ambiguity, since A must be of the same affection as a. It will be seen from these formula that there are limitations of the data in order to insure a possible 'triangle; in fact, c must lie between a and s-a in order that the values found for cos b, cos B, and sin A may be 'numerically not greater than unity. If c and a are right angles, A is a right angle, and b and B aro indeterminate. 1 76. Ilaring given the two angles A and B. Here we have from (5) and (6) of Art. 62, COS c = cot A cot B, COS (1 COS A sin /i's cOS = cos B sind. Here c, a, b are determined without ambiguity. There are limitations of the data in order to insure a possible triangle First suppose A less than then B must lie between 2' - A 394 + 1; 2 TT next suppose A greater than then B must lie between ) * T 2-(-A) and 3+ (- - A), that is, botween A 37 and - A. 2 77. Having given a sile a and the opposite angle A. Here we have from (2), (4) und (6) of Art. 62, sin a COS sin sin b= tan a cot A, sin B = sin A' COS a Here there is an ambiguity, as the parts are determined from their sines. If sin a be less than sin A, there are two values admissible for c; corresponding to each of these there will be in general only one admissible.value of b, since we must have COS C = cos a cos 6, and only one admissible value of B, since we must have cos c = cot A cot B. Thus if one triangle exists with the given parts, there will be in general two, and only two, triangles with the given parts. We say in general in the preceding sentences, because if a = A there will be only one triangle, unless a ·SOLUTION OF RIGHT-ANGLED TRIANGLES. 43 and A are each right angles, and then 6 and B become indeterininate. It is casy to see from a figure that the ambiguity must occur in general. B C For, suppose BAC to be a triangle which satisfies the given conclitions; procluce AB and AC to ineet again at A'; then the triangle .I' BC also satisfies the given conditions, for it has a right angle at C, BC the given side, and A' = A the given angle. If, a A, then the formulze of solution shew that c, b, and B are right angles; in this case A is the pole of BC, and the triangle A'BC is symmetrically equal to the triangle ABC (Art. 57). If a and I are both right angles, B is the pole of AC; B and b are then equal, but may lave any value whatever. There are limitations of the data in order to insure a possible trianglo. A and a must have the same affection by Art. 64; hence the formulæ of solution shew that a must be less than A if both are acute, and greater than d if both are obtuse. EXAMPLES. a cos If ABC be a triangle in which the angle C is a right angle, prove the following relations contained in Examples 1 to 5 6 6 1.' Sinº sin + cossins 2 2 6 tan 2. Tan (+a) tan dį (c-a) = tanº 2 A 3. Sin (c - b) = tan sin (c +b). 0 ·EXAMPLES. . 4. Sin a tan A – sin 8 tấn B= sin (a - b). 5. Sin (c-a) = sin b cos a tan į B, . Sin (c-a)= tan b cos c tan & B. 6. If ABC be a spherical triangle, right-angled at C, and cos A = cos a, shew that if A be not a rigl angle b+c=&or 3 7, according as b and & are both less or both greater than 2 TO C +7. If a, ß be the arcs drawn from the right angle respectively perpendicular to and bisecting the lıypotenuse c, shew that siuº (1 + sinº a) = sinº B. 8. In a triangle, if C be a right angle and D the middle point of AB, shew that 4 cos min CD= sin' a + sin 6. cosami 9. In a right-angled triangle, if 8 be the length of the arc drawn from C perpendicular to the hypotenuse AB, show that cot 8= (cot a +cot' 8). 10. 044, is a spherical triangle right-anglod at 4, and acuteangled at A; the arc A,4, of a great circle is drawn perpendicular to 04, 4, 4, is drawn perpendicular to 041, and so on: shew that 4 Ant, vanishes when n becomes infinite; and find the value of cos AA, cos 4,4,cos A 4.......to infinity. 11. ABC is a right-angled spherical triangle, A not being the right angle: shew that if A = a, then c and b are quadrants. 12. If & be the length of the arc drawn from C perpendicular to AB in any triangle, shew that cos d=cosec c (cos' a + cos 6 - 2 cos a cos 6 cos os c). 13. ABC is a great circle of a sphere; A A', BB', CC", are avcs of great ciroles drawn at right angles to ABC and reckoned posi- ·EXAMPLES. tive when they lie on the same side of it: shew that the condition of A', B', C' lying in a great circle is tan AA' sin BC + tan BB-sin CA + tan CC'sin AB=0. 14. Perpendiculars are drawn from the angles 4, B, C of any triangle meeting the opposite sides at D, E, F respectively: shew that tan BD tan CE tan AF=tan DC tan E A tan FB. 15. 0.c, Oy are two great circles of a sphere at right angles to each other, P is any point in AB another great circle. OC=p is the arc perpendicular to AB from 0, making the angle COx = a with Ox. PJ, PN are arcs perpendicular to Ox, Oy respectively: shew that if OM = x and ON=y, cos a tan x + sin a tan y=tan p. 16. The position of a point on a sphere, with reference to two great circles at right angles to each other as axes, is determined by the portions 0, $ of these circles cut off by great circles through the point, and through two points on the axes, each from their point of intersection ! shew, that if the three points (0,4), (6,4), (0', $) lie on the same great circle tan $ (tan 6' – tan 0") + tan so' (tan 8'"– tan ) + tan $" (tan 8 - tan 6')= 0. 17. If a point on a sphere be referred to two great circles at right angles to each other as axes, by means of the portions of these axes cut off by great circles drawn through the point and two points on the axes each 90° from their intersection, shew that the equation to a great circle is tan A cot a + tan o cot ß= 1. T B=3 and C= shew. C= 7 2 18. In a spherical triangle, if A = 5, B = thout a +b+c=1 ·( 46 ) • VI. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 78. The solution of oblique-angled triangles may be made in some cases to depend immediately on the solution of right-angled triangles; we will indicate these cases before considering the subject generally. (1) Suppose a triangle to have one of its given si les equal to a quadrant. In this case the polar triangle has its corresponding angle a right angle; the polar triangle can therefore be solved by the rules of the preceding Chapter, and thus the elements of the primitive triangle become known. (2) Suppose among the given elements of a triangle there are two equal sides or two equal angles. By drawing an arc from the vertex to the middle poin of the buse, the triangle is divided into two equal right-angled triangles; by the solution of one of these right-angled triangles the required elements can be found. (3) Suppose among the given clements of a triangle there are two sides, one of which is the supplement of the other, or two angles, one of which is the supplement of the other. Suppose, for example, that b+c=n, or else that B+C ==; produce BA and BC to meet at B (see the first figure to Art. 38); then the triangle B'AC has two equal sides givert, or else two equal angles given; and by the preceding case the solution of it can be made to depenál on the solution of a right-angled triangle. 79. We now proceed to the solution of oblique-angled triangles in general. There will be six cases to consider. COS a COS CON C 80. Having given the three sides. b Here we have cos A and similar formulæ sin b sin c frog.cos B and cos C. Or if we wish to use formulæ suited to loga- 9 ·SOLUTION OF OPLIQUE-ANGLED TRIANGLES. rithms, we may take the formula for the sine, cosine, or tangent of half an angle given in Art. 45. In selecting a formula, attention should be paid to the remarks in Plane Trigonometry, Chap. XII. towards the end. 81. Ilaving given the three angles. cos + cos B cos C Here we have cos a and similar formula sin B sin C for cos b and cos c. Or if we wish to use formulæ suited to logarithms, we may take the formula for the sine, cosine, or tangent of lalf a side given in Art. 49. There is no ambiguity in the two preceding cases; the triangles however may be impossible with the given elements. > 82. Jaring yiven two sides and the included angle (a, C, b). By Napier's analogies cos! (a - b) tan? (A + B) = cott C, cos & (a + b) sin) (a-6) tan d (01-B) cotC; , sin (a + b) these determine 3 (A + B) and ] (4-B), and thence A and B. sin a sin C Then c may be found from the formula sin c= ; in sin A this case, since c is found from its sine, it inay be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater side of a triangle is opposite to the greater angle. Or we may determine c from equation (1) of Art. 54, which is free from ambiguity. Or we may determine c, without previously determining A and B, from the formula cos c = cos a cos b + sin a sin b cos C; this is free from ambiguity. This formula may be adapted to logarithms thus: cos c = cos 0 (cos a + sin a tan 6 cos C);

·48 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. Assume tari 0 = tan 6 cos Cs then cos 6 cos (a - A) OOS c = cos 6 (cos a + sin a tan 6) cos o this is adapted to logarithms. . D B B Or we may treat this case conveuiently by resolving the triangle into the sum or difference of two right-angle triangles. From A draw the arc AD perpendicufar to CB or CB produced; then, by Art. 62, tan CD = tan b cos C, and this determines CD, and then DB is known. Again, by Art. 62, cos 6 COS C = COS AD cos DB = COS DB Cos CD; this finds c. It is obvious that CD is what was denoted by 0 in the former part of the Article. > By Art. 62, tan AD = tan C sin CD, and tan AD = tan ABD sin DB; thus tan ABD sin DB = tan C sin , where DB= a -- or 6-a, according as D is on CB or CB produced, and ABD is either B or the supplement of B; this formula enables us to find B independently of A, Thus, in the present case, there is no real ambiguity, and the triangle is always possible.

·tan c, SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 49 83. Ilaving given two angles and The included side (A, C, B). By Napier's analogies, tan (a + b) = cos (A - B) cos (A + B) sin } (A - B) tan } (a - b) = tanc; sin (A + B) these determine } (a + b) and } (a,0), and thence a and b. sin A sin c Then C may be found from the formula sin C: ; in sin a this case, since C is found from its sine, it may be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater angle of a triangle is opposite to the greater side. Or we may determine C from equation (3) of Art. 51, which is free from anıbiguity. Or we may determine C without previously determining a and b from the formula cos C = cos A cas B + Sin A sin B cos C. This formula anay be adapted to logarithms, thus: cos C = cos B (- cos A + sin A tan B cos c); assume coto= tan B cosc; then cos B sin (A -) = cos B(- cos A + cot $ sin A) this is adapted to logarithms. Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to CB (see the rightland figure of Art. 82); then, by Art. 62, cos c = cot B cot DAB, aud this determines DAB, and then CAD is known. Again, by Art. 62, cos AD sin CAD = cos C and cos AD sin BAD = cos B; cos C ; sin ¢ therefore cos C sin CAD cos B sin BAD ; this finds C. It is obvious that DAB is what was denoted by $ in the former part of the Article. T. S. T.

·50 SOLUTION OF OBLIQUE-ANGLED TRIANGLES, By Art. 62, tan AD-tan AC cos CAD, and tan AD = tan A B cos D.ID; thus tan b cos CAD = tan c cos , where CAD=A-; this formula enables us to find b independently of a. Similarly we may proceed when the perpendicular AD fulls on CB produced; (see the left-hand figure of Art. 82).' Thus, in the present case, there is no real ambiguity; moreover the triangle is always possible. 84. Having given two sides and the angle opposite one of them (a, b, A). The angle B may be found from the formula sin 6 sin B sin A, sin a and then C and c from Napier's analogies, tan C cos } (a - b) cos (a + b) cot (A + B), cos }(A + B) tan c = tan } (a + b). cos 1(A -B) In this case, since B is found from its sine, there will sometimes be two solutions, and sometimes there will be no solution at all, namely, when the value found for sin B is greater than unity. We will presently return to this point. (See Art. 86.) We may also determine C and c independently of B by formulæ ai ed to logarithms. For, by Art. 44, cot A cota sin b = cos 6 cos C + sin C cot A=cos 6 (cos C + sin C); cot A assume tan $ = cos 6 cos b cos (C-0) cot a sin b = cos 6 (cos C + tan $ sin C)=- cos o therefore Cos (0-4)= cos do cot a tands cos 6 ; thus ·SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 51 COS a from this equation C-% is to be found, and then C. The ambiyuity still exists; for if the last equation leads to C-6=Q, it will be satisfied also by 0-C=a; so that we have two admissible values for C, if ø+a is less than and $-a is positive. And cos 6 cos c+sin sin c cos A = cos b (cosc + sin c tau: 3 cos 4); assame tan 6 = tan 6 cos A; ; thus 5 cos pc - 0) = cos 7 (cos c+sin c tan ) = cos a cos therefore cos (c-6) from this equation c A is to be found, and then c; and there may be an anbiguity as before. Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles. COS COS a ; cos ; cos 6 D B' Let CA = b, and let CAE= the given angle A ; from C draw CD perpendicular to AE, and let CB and CB' =a; thus the figure shews that there may be two triangles which have the given elements. Then, by Art. 62, cos b = cot A cot ACD; this finds ACD. Again, by Art 62, tan CD=tan AC cos ACD, and tan CD= tan CB cos BCD, or tan CB' cos B'CD, therefore tan AC COS ACD= tan CB cos BCD, or tan CB cos BºCD; this finds BCD or B'CD. It is obvious that ACD is what was denoted by p in the former part of the Article. E 2 ·52 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. Also, by årt. 62, tan AD'= tan AC cos A; this finds AD. Then = cos CD cos AD, = cos CD cos BD, cos CB = cos CD cos B'D; cos ACcos CB or therefore COS AC COS AD cos CB cos BD or cos CB cos ¿Di this finds BD or BD. It is obvious that AD is what was denoted by 0 in the former part of the Article. 85. Having given two angles and the side opposite one of them (A, B, a). This case is analogous to that immediately preceding, and gives rise to the same ambiguities. The side b may be found froin the formula sin B sin a sin b sin 4 and then C and c from Napier's analogies, cos } (a - b) tan C cot (4+B), cos } (a + b) cos 4 (A + B) tan ļ (a + b). cos (A - B) tan c - We may also determine C and c independently of b by formula adapted to logarithms. For cos A =-cos B cos C + sin B sin C cos a = cos B (- cos C + tan B sin C cos a), Assume cot $ = tan B cos a; thus = cos B (- cos C + sin C çot ) = COS B sin(C–os), cos 4: sinº ; therefore cos A sin $ sin (C-4) cos B ·SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 53 - from this equation C $ is to be found and then C. Since C-8 is found from its sine there may be an ambiguity. Again, by Art. 44, cot a sin cot Asin B =cota sin c cos c cos B=cos B COS C + 2), cos B cot a assume coto ; then cos B cos B sin (c-) cot A sin B =cos B(- cos c+ sin c cot ) = sin @ therefore sin (c - ) = cot A tan B sin 0; from this equation c O is to be found, and then c. Sincec A is found from its sine there may be an ambiguity. As before, it may be shewn that these results agree with those obtained by resolving the triangle into two right-angled triangles; for if in the triangle ACB the arc CD be drawn perpendicular to AB', then BCD will = 4, and B'D = 0. 86., We now return to the consideration of the ambiguity which may occur in the case of Art. 84, when two sides are given and the angle opposite one of them. The discussion is somewhat tedious from its length, but presents no difficulty. Before considering the problem generally, we will take the particular case in which a=b; then A must B. The first and third of Napier's analogies give cot C = tan A cos a, tan c= tan a cos'A; now cot šC and tan &c must both be positive, so that A and a must be of the same affection. Hence, when a=b, there will be no solution at all, unless A and a are of the same affection, and then there will be only one solution ; except when A and a are both right angles, and then cot C and tan c are indeterminate, and there is an infinite nuinber of solutions. We now proceed to the general discussion. If sin b sin A be greater than sin a, there is no triangle which satisfies the given conditions ; if sin b sin A is not greater than ·54 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. sin o sin A sin a, the equation sin B = furnishes two values of B, sin a which we will denote by B and B', so that p = -77-B; we will suppose that B is the one which is not greuter than the other. Now, in order that these values of B may be' aulmissible, it is necessary and sufficient that the values of cot ļC and of tan de should both be positive, that is, A - B and a-6 must have the same sign by the second and fourth of Napier's analogies. We have therefore to compare the sign of 1-B anıl the sign of A-B' with that of a-o. We will suppose that A is less than a right angle, and separate the corresponding discussion into three cases. 7T I. Let 6 be less than 22 sin 6 (1) Let a be less than b; the formula sin B - sin al makes sul a B greater than d, and à fortiori B' greater than 1. Hence there are two solutions. (3) Let a be equal to b; then there is one solution, as porta viously shewn. (3) Let a be greater than b; we may have then a + b less than T or equal to 7 or greater than 7. If a+b is less than TT, then sin a is greater than sin b; thus ß is less than A and therefo. c admissible, and ß' is greater than A and inadmissible. Hence there is one solution. If a + b is equal to it, then ß is equal to A, and B' greater than A, and both are inadmissible. Hence there is no solution. If q + b is greater than 7, then sin a is less than sin b, and B and B are both greater than A, and both inadmissible. Hence there is no solution. II. Let 6 be equal to 2 (1) Let a be less than b; then B and B' are both greater than A, and both admissible. Hence there are two solutions. (2) Let a be equal' to b; then there is no solution, as previously shewn.

·SOLUTION OF OBLIQUE-ANGLED TRIANGLES: 55 (3) Let a be greater than b; then sin a is less than sin b, and B and B' are both greater than A, and inadmissible. Hence there is no solution. III. Let 6 be greater than 2 (1) Let a be less than b; we may haye then a + b less than or equal to 7 or greater than 7. If a +b is less than 7, then sin a is less than sin b, and B and ß are both greater than A and both admissible. Hence there are two solutions. If a +b is equal to 7, then ß is equal to A and inadmissible, and B is greater than A and admissible. Hence there is one solution. If a + b is greater than 7, then sin a is greater than sin ; B is less than A and inadmissible, and p' is greater than A and admissible, llence there is one solution. (2) Let a be equal to b; then there is no solution, as previously shitwn. (3) Let a be greater than b; then sin a is less than sin b, inil B and B' are both greater than 1 and both inadmissible. Hence there is no solution. We have then the following results when A is less than a right angle. two solutions, 6 ...one solution, a>b and a + b <T one solution, a>b and a + b no solution. a<b TT < T OL' >7T a< 영 ...... two solutions, .no solution. O a=b or a > 0 b> a <b and a + b <T asb and a +b = Or > T a=b or > >b ...... two solutions, one solution, .no solution, ......

·56 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. It must be remembered, however, that in the cases in which two solutions are indicatod, there will be no solution at all if sin a be less than sin o sin A. In the same manner the cases in which A is equal to a right angle or greater than a right ungle may be diſcussed, and the following results obtained. When A is equal tova right angle, a<b or a= :6 a>b and a + b < a>b and a+b=ī or > NI .no solution, ...one solution, .no solution. be ſa<b or a > ... asb and a + b > asb and a+b=ī or <T a=b or ab .....no solution, .infinite number of solutions. ...one solution, ..no solution, .no solution. b = 1/2 When A is greater than a right angle, asb or a=b b a>b and a + b =or <T lamb and a + b > (a<b or a=b 2 lab..... rasb and a +b> asb and a + b = =т оr <т 6> a=b ..... la=b ..... . .no solution, one solution, two solutions. .no solution, two solutions. one solution, .no solution, ...one solution, .two solutions. ... As before in the canes in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A. It will be seen from the above investigations that if a lies between 6 and r -6, there will be one solution ; if a does not lie between b and -b either there are two solutions or there is no solution; this enunciation is not meant to include the . whigh a=b.or =r-6. cases in ·SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 57 87. The results of the preceding Article may be illustrated by a figure. Let ADAE be a great circle; suppose P4 and PA' the projections on the plane of this circle of arcs which are each equal.to 7 and inclined at an angle A to ADA'; let PD and PE be the projections of the least and greatest distances of P from the great circle (see Art. 59). Thus the figure supposes and b each less than 2 If a be less than the arc which is represented by PD there is no triangle; if a be between PD and PA in magnitude, there are two triungles, since B will fall on ADA', and we have two triangles BPA and BPA'; if a be between PA and PH there will be only one triangle, as B will fall on A'H or AH', and the triangle will be either APB with B between A' and H, or else A'PB with B between A and II', but these two triangles are symmetrically equal (Art. 57); if a be greater than PH there will be no triangle, The figure will easily serve for all the cases; thus if 4 is greater than we can suppose PAE and PA'E to be equal to A; if 2 T TT b is greater than we can take PH and PH' to represent b.

·58 • SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 88. The ambiguities which occur in the last case in the solution of oblique-angled triangles (Art. 85) may be discussed in the same manner as those in Art. 86; or by nyans of the polar triangle, the last case may be deduced from that of Art. 86. EXAMPLES. 1. The sides of a triangle are 105°, 90", and 73° respectively: find the sines of all the angles. Sills sin (s-c) 2. Shew that tand tan B Sulve a triangle when a side, an adjacent angle, and the sum of the other two sides are given. 3. Solve a triangle kaving given a side, an avljacent angle, and the sum of the other two augles. 4. A triangle has the sum of two sides equal to a semicircumference: find the arc joining tho vertex with the middle of the base. 5. If a, b, c are known, c being a quadrant, deterrnine the angles: shew also that if & be the perpendicular ou c froin the opposite angle, cos' = cosa + cos'b. 6. If one side of a spherical triangle be divided into four equal parts, and 0,, 0, 0,5, 0, be the angles subtended at the opposite angle by the parts taken in order, shew that sin (0,+) sin 0, sin 0,= sin(0, +0.) sin 6, sin ,. 7. In a spherical triangle if 4 =B=2C, shew that 3 ain (a + ) ingco. - eina ·CIRCUMSCRIBED AND INSCRIBED CIRCLES. 59 8. In a spherical triangle if A=B= 2C, shew that C COS 2 8 sin" (cos & s+ sina) 1. CON a AG = sin sec 9. If the equal sides of an isosceles triangle ABC be bisected by an arc DE, and BC be the base, shew that DE BC sin 2 2 10. If ,, C, be the two values of the third side when A, a, 6 Are given and the triangle is ambiguous, shew that tauri tan = t:n } (6 – a) tan $ (6+ a). VII. CIRCUMSCRIBED AND INSCRIBED CIRCLES. 89. To find the angular radius of the small circle inscribeu in a given triungle. Let ABC be the triangle; biscct the angles A and B by arcs meeting at P; from P draw PD, PE, PP perpendicular to the sides. Then it may be shewn that PD, PE, PF are all equal; also that. A E = AF, DI'= BD, CD=CE. Hence BC + AF = half the sum of the sides = 8; therefore AF=8-a. Let PF=r. Now tan PF = tan PAF sin AF (Art. 62); thus tanr = tan sin (8-a) .(1). 2 ·60 CIRCUMSCRIBED AND INSCRIBED CIRCLES. The value of tan , may be expressed in various formis ; thus from Art. 45, we obtain (sin (8 – ) sin (8-6 tan sin s sin (8 - «) 1-1 Vasant substitute this value in (1), thus sin (8-a) sin (8-6) sin (8-c); n tanr = {{sin (e- :-)) 2. (Art. 46)....(?). SI 8 SU 8 Again sin (8 – a) = sin {} (b+c) - fa} = sin(6 + c) cosa cos } (b + c) sind a sin a cos} a {cos $ (B-C) - cos } (B+C), (Art, 5+) sin A sin a sin), B sinf sin A sin B sin C therefore from (1) tan r = sin d (3); i cost hence, by Art. 51; tanr= Vf-cos S cos (- A) cos (S – B) cos (S --C)} 2 cos A cus į B cosa N 2 cos s A cosB cos (+). It may be shewn by common trigonometrical formule that 4 cos 4 cosB coB 4C = cos S + cos (S-A) + cos (S – B) + cos (S--C); hence we have from (4) cotg = 1 2N cos S + cos (S - A) + cos (S – B) + cos (S09}....(5).

·CIRCUMSCRIBED AND INSCRIBED CIRCLES. 61 90. To find the angular radius of the small circle described so as to touch one side of a given triangle, and the other sides produced. B A C Let ABC be the triangle; and suppose we require the radius of the small circle which touches BC, and AB and AC produced. Produce AB and AC to meet at A'; then we require the radius of the small circle inscribed in A'BC, and the sides of A'BC are a, #-6, - respectively. Hence if r, be the required radius, and 8 denoté as usual } (a +b+c), we have from Art. 89, tan r, 5 tan sin s 2 ..... (1). n From this result we may derive other equivalent forms as in the preceding Article; or ye may make use of those forms immediately, observing that the angles of the triangle A'BC are A, 7-B, 7-C respectively. Hence & being fa+b+c) and being ] (A + B + C) we shall obtain sin s sin (8-6) sin (8 - c) ta ...(2), sin (s-a) sin (s-a) cos B cosc sin a (3), cosA {-cos S cos (S- A) cos (S – B) cos (S-C)} 2 cos A sin B sino N (4), 2 coss A sin B sin 0 $ tan r, ........... tan, cott, 1 2N V - cos S - cos (S – A) + cos (S-B) + cos (S (S-C) ...(5).

·62 CIRCUMSCRIBED AND INSCRIBED CIRCLES. These results may also be found independently by bisecting two of the angles of the triangle A'BC, so as to determine the pole of the small circle, and proceeding as in Ast. 89. 91. A circle which touches one side of a triangle and the other sides produced is called an escribed circle, thus there are three escribed circles belonging to a given triangle. We may denote the radü of the escribed circles which touch CA and AB respectively by., and rs, and values of tan r, and tanr, may be found from what has been already given with respect to tan, by appropriate changes in the letters which denote the sides and angles. In the preceding Article a triangle A'BC was formed by producing AB and AC to meet again at A'; similarly another triangle may be formed by producing BC and BA to meet again, and another by producing CA and GB to meet again. The original triangle ABC and the three formed from it have been called associated triangles, ABC being the fundamental trian'le. Thus the inscribed and escribed circles of a given triangle are the same as the circles inscribed in the system of associated triangles of which the given triangle is the fundament el triangle. 92. To find the angular radius of the small circle described about a given triangle. PI E D Let ABC be the given triangle; bisect the sides CB, CA at and respectively, and draw from D and E arcs at right angles te OB and CA respectively, and let P be the intersection of these ·CIRCUMSCRIBED AND INSCRIBED CIRCLES. 63 arcs. - ......... cos S cos S Then P will be the pole of the small circle described about ABC. For draw PA, PB, PC; then from the right-angled triangles PCD and PBD it follows that PB=PC; and from the right-angled triangles PCE and PAE it follows that PA= PC; hence PA = PB.= PC. Also the angle PAB = the angle PBA, the angle PBC = the angle PCB, and the angle PCA = the angle PAC; therefore PCB + A = } (4 +B+C), and PCB=S-A. Let PC =R. Now tan CD= tan CP cos PCD, (Art. 62), thus tanda = tan R cos (S - A), tanţa therefore tan R (1). cos (8 - A) The value of tan R may be expressed in various forms; thus if we substitute for tan from Art. 49, we obtain 2 55 tan R. (2). (cos (S – A) cos (8 – B) cos (S – N C Again cos (S – A)+cos {} (B+C)-4 A} =cos I (B+C) cos. A + sin }(B+C) sind A sin A cos f A {cos } (b + c) + cos (6-c)}, (Art. 54,) costa sin A cos fb cos c; cos sa therefore from (1) sin fa (3). sin A cos fb cos Substitute in the last expression the value of sin A from Art. 46; thus 2 sina sin sinc tan R J{sin 8 sin (8-a) sin (8-0) sin (8 -c)} 2 sin f a sin £ b sino (4) { tan R c .......

·4 6+ CIRCUMSCRIBED AND INSCRIBED CIRCLES. It may be shown, by corfimon trigonometrical formulæ, that 4 sin f a sin fb sin $ c = sin (8 – a) + sin (8 – 6) + sin (8 - c) – sin 8; hence we have from (4) 1 tan R sin (8-a) + sin (8 - 6) + sin (s - c) – sin 2n (5). - sin o) 93. To find the angular radii of the small circles described round the triangles associated with a given fundamental triangle. Let R, denote the radius of the circle described round the triangle formed by producing AB and AC to meet again at A'; similarly let R, and R, denote the radii of the circles described round the other two triangles which are similarly forined. Then we may deduce expressions for tan R,, tan R,, and tan R, from those found in Art. 92 for tan R. The sides of the triangle A'BC are a, 1-6, 7-C, and its, angies are A, 1-B, -C; hence if 8 = $ (a + b + c) and S = {(A + B + C') we shall obtain from Art. 92 10 tan ka tan R (1), 1 cos S 0 tan R, /{cus cos (S - A) N (?), ... tan R, cos (S - A) Cox S cos ($- b) cos (5 – C) sin fa sin A sin o sin c 2 sin fa cost cosc {sin 8 sin (8 – a) sin (8-6) sin (8 -- c)} (3), ....... tan R, .(4), tan R in { ain s– sín (8 – a) + sin (8–6) + sin (8–c)}...... (5). 2n Similarly we may find expressions for tan R, and tan Rg. 94. Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles. We will give one that will be of use hereafter.

·CIRCUMSCRIBED 'AND INSCRIBED CIRCLES. 65 To prove that (cotr + tan R) (sin a + sin b + sin c)'- 1. 2 40 We have 4n=1-cosa - cos' 6 - cos c+ 2 cos a cos 6 cos c; ; therefore (sin a + sin b + sin c). – 413 = 2(1 + sin a sin b + sin b sin c + sin c sin a - cos a cos 6 cos se). 1 Also cotr + tan R = (sin 8 + sin (8-a)+sin (846) + sin (s in (8–c)}; 2r21 and by squaring both members of this equation the required result will be obtained. For it may be shewn by reduction that sinº 8 + sin" (8 – a) + sinº (8 — b) + sinº (8 + C) = 2 – 2 cos a cos 6 cos C, and sin s sin ($ – a) + sin & sin (s-6) + sin s sin (8 - c) + sin (8 - a) sin (8-6)* sin (8 — 6) sin (8-c) + sin (8—c) sin (8- a) = sin a sip b+ sin b sin c + sin c sin a. Similarly we may prove that 1 (cot r, - tan R) (sin 6 + sin c - sin a) – 1. 4n" 7T 95. In the figure to Art. 89, suppose DP produced through P to a point A' such that DA' is a quadrant, then is a pole of BC, and PA 2 -r; similarly, suppose EP produced through P to a point 'B' such that EB' is a quadrant, and FP produced through P to a point c' such that FC" is a quadrant. Then ABC is the polar triangle of ABC, and PA' = PB'= PC": 2 Thus P is the pole of the small circle described round the polar triangle, and the angular radius of the small circle described round the polar triangle is the complement of the angular radius of the T. S. T, ·66 CIRCUMSCRIBED AND INSCRIBED CIRCLES. small circle inscribed in the primitive triangle. And in like manner the point which is the pole of the small circle inscribed in the polar triangle is also the pole of the sipall circle described round the primitive triangle, and the angular radii of the two circles are complementary. EXAMPLES. 1 19.30 + 4, In the followiniy examples the notation of the Chapter is retained. Shew that in any triangle the following relations bold contained in Examples 1 to 7: 1. Tan TI tan r, tan , tanr sin s. 2. Tan R+ cot r = tan R, + cot r = tan R, + cotr, tan Ng + cotr, = (cot r + cot ri + cot r, + cotrs). 3. Tan® R+ tan'R, + tanR, + tanº R, cotºr + cot' r, + cotr, + cot? Tanr, + tan r, + tan r, - tan r (1 + cosa + cos b + cos c). cotr, + cot r, + cotr, - cotř 5. Cosecr4cot(8-a)cot(8-5)+cot(8-6)cot(3-c)+cot (8-c)(x-a). 6. Cosec* r, =cot (8-6) cot(s —c)-cot & cot (8-0)-cots cot (8-c). 7. Tan R, tan R, tan R, = tan R sec S. Shew that in an equilateral triangle tan R=2 tanr. 9. If ABC be an equilateral spherical triangle, P the pole of the circle circumscribing it, Q any point on the sphere, shew that cos (A + cos QB+cos QC = 3 cos PA cos PQ. 10. If three small circles be inscribed in a spherical triangle having each of its angles 120°, so that each touches the other two as well as two sides of the triangle, show that the radius of each of the small circles = 30', and that the centres of the three small sircies coincide with the angular points of the polar triangle. متر ·( 67 ) VIII. AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 96. To find the area of a Lune. A Lune is that portion of the surfice of a sphere which is comprised between two great semicircles. (DEF B Let ACDDA, ADD' be two lunes having equal angles at A ADD then oue of these lunes may be supposed placed on the other so as to coincide exactly with it; thus lunes having equal angles are cqual. Then by a process similar to that'used in the first proposition of the Sixth Book of Euclid it may be shewn that lunes are proportional to their angles. Hence since the whole surface of a sphere may be considered as a lune with an angle equal to four right angles, we have for a lune with an angle of which the circular measure is A, area of lune A surface of sphere27 Suppose r the radius of the sphere, then the surface is 4arget (Integral Calculus, Chap. VII.); thus area of lune = 274779 = 24/?. F2

·68 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS.
297. To find the area of a Spherical Triangle.
.
B
Let ABC be a spherical triangle; produce the arcs wluch forin
its sides until they meet agaiu two and two, which will happen
when each has become equal to the semi-circumference. The
triangle ABC now forms a part of three lunes, namely, A BDCA,
BCEAB, and CAFBC. Now the triangles CDE and / FB are
subtended by vertically opposite solid angles at 0, and we will
assume that their areas are equal; therefore the lune CAFBC is
equal to the sum of the two triangles ABC and CDE. Hence if
A, B, C denote the circular measures of the angles of the triangle.
we have
triangle ABC + BGDC = lune ABDCA = 247",
triangle A BC + AHEC= lune BCEAB = 2 Br",
triangle ABC + triangle CDE=lune CAFBC = 2Cpl ;
hence, by addition,
twice triangle ABC + surface of hemisphere = 2 (4 + B + C) pe;
tberefore
triangle ABC = (4 + B+C - +) .
The expression A+B+C - is called the spherical excess of
the triangle; and since
.
A+B+C - 2*r",
(4 + B + C - 7) gol
27

·AREA OF A SPHERICAL TRIANGLE. TRIANGLE SPHERICAL EXCESS. 69 . the result obtained may be thus enunciated: the area of a spherical triangle is the same fraction of half the surface of the sphere as the spherical excess is of four right angles. 98. We have assumed, as is usually done, that the areas of the triangles CDE and AFB in the preceding Article are equal. The triangles are, however, not absolutely equal, but symmetrically cqual (Art. 57), so that one cannot be made to coincide with the other by superposition. It is, however, easy to decompose two such triangles into pieces which admit of superposition, and thus to prove that their areas are equal, For describe a small circle round each, then the angular radii of. these circles will be equal by Art. 92. If the pole of the circumscribing circle falls inside each triangle, then each triangle is the sum of three isosceles triangles, and if the pole falls outside each triangle, then ench triangle is the excess of two isosceles triangles over a third ; anıl in cach case the isosceles triangles 'of one set are respectively absolutely aval to the corresponding isosceles triangles of the other set. . 99. To find the anca of a spherical polygon. Let n be the number of sides of the polygon, the sum of all its angles. Take any point within the polygon•and join it with all the angular points ; thus the figure is divided into n triangles. Hence, by Art. 97, area of polygon = (sum of the angles of the triangles – Nat)r", and the sum of the angles of the triangles is equal to together with the four right angles which are formed round the common vertex; therefore area of polygon = {s- (m- 3- (n − 2)}-2. This expression is true even when the polygon has some of its angles greater than two right angles, provided it can be decomposed into triangles, of which each of the angles is less than two right angles.

·70 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 100. We shall now give some expressions for certain trigonometrical functions of the spherical excess of a triangle. We denote the spherical excess by E, so that E= A + B + C - 18. . € 101. Cagnoli: Theorem. To shew that sin } E = /{sin 8 sin (8 – a) sin (8 – b) sin (8 –c)} 2 cos a cost costa Sin $ E=sin }, /4 +B+C – 7) = sin {} (x1+B) – 1 (1 - C)} = sin } (+B) sin 1 C - cos } (A + B) cos 1C sin C cos! {cos (a - b) - cos (a + b)}, (Art. 54), 'cosc sin C'sid } a sin 1 6 cosc sin } a sin b 2 sin 8 sin (8-a) sin(8-7) sip (8 - c)} cos C sin a sin b. J{sin e sin (8-a) sin (8-6) sin(8-0)} 2 cosa cos vos • M C 102. Lhuilier's Theorem. To shew thut tan 1E=}{tan 1 8 tan } (8-a) tan 1 (8 - 6) tan! (8-c)}. sin 1 (A+B+C -) Tan Ecos (A+B+C) sin (A+B) - sin 1 (T-C) (Plane Trig. Art. 84), cos (A + B)+cos (-C)' sin 1 (A + B) - cos C cos: (4 +B) + sinu cos 1 (a - b) - cos 1c cos1 C (Art. 54). cos : (a + b) + cos c. sin ? Hence, by Art. 45, we obtain tan E = sin 8 sin (8-0) sin 4 (c+a-b) sin 1 (c+b-a) cos 4 (a+b+c) cos * (a+b-c) sin (8 - a) sin (8 - 6) =1/{tan 1 8 tan 1 (8 - a) tan : (8-6) tan 1 (8 - c)}. + ·AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 71 103. We may obtain many other formula involving trigonometrical functions of the spherical excess. Thus, for example, cos 1 E = cos {I (4+B)-1 (-C)} = cos (4+B) sin 1C + sin ) (A + B) cos 10 = 605 ! (a+b) sanº} C + cos ! (@–6) cos"}C} sce£e, (Art.54), - coa + sin ţa sin { / (cos" } 0 - sinº ļC)} scc ļo cos. u cos (cos' C + sin C) = cos a cos } b + sin a sin } b cos C} sec. ..... (1). Agruin, it was shown in Art. 101, that sin ! E = sin C sin } & sin 16 secc; sin } a sin i b sin C therefore tan Ecos ), a cos } ] + sin } a sin } cos C (2). Again, we have from above cos. E = cos } a cos } 6 + sìn ļ a sin fb cos c}sec ļa C (1 + cos a)(1 + cos 6) + sin a sin b cos C 4 cos 1 a cos 16 cos :: 1 + cos a + cos b + cos C cos 1 a + cos' } b + cos' } c -1 4 cosa cos } b cosc 2 cos a cos } b cosc -. (3). In (3) put 1 - 2 sin' E for cos 4 E ; thus sin E- 1 + 2 cos 1 a cos } 6 cos 1c cos"}a - sos' } 6 – cos® ]c. 4 cost a cos 1b cosc + By ordinary development we can shew that the numerator of the above fraction is equal to 4 sin | 8 sin } (8 - a) sin 1 (8-6) sin 1 (8-c);

·72 AREA OF A SPHERICAL TRIANGLE SPHERICAL EXCESS. ...(4) therefore sin 1 8 sin (8-a) sin 1 (8 - b) sin } (8 - sinº 1 E cos da cos } b cosec Similarly cos' | E cos 1 8 cos 1 (8 – a) cos } (8-6) cos } (x-c) cosa cos Ib cosc Hence by division we obtain Lhuilier's Theorem. .(5). Again, sin (C-1E) sin C cot E - cos C sin E cos) a cos db + sin } a sin 6 cos C = sin c cos C, by (2), sin a sino sin C cot 1 a cot1b; therefore, by Art. 101, sin (C-1) = {{sin 8 sin (: - a) sin (8-7) sin(s - c)} 2 sin } a sin } b cos,c Again, cos (C-1E)= cos C cos! E + sin C sin } E (1+cosa)(1+cosb) cosC+sin asin b cos' C +sinºC sin a siniscelo 4 cos a cos 16 cos c (1 + cos a)(1 + cos 6) cos C + sin a sin 6 4 cos a cos 1 b cos C ={cos ļ a cos } cos C + sin ţ a sin {B} sec o COS C sin a sin b cos C + 4 sin" } a sin' 7 6 4 sina sin fb cosc cos a cos 6 + (1 cos a)(1 -cos) 4 sin f a sin fb cosc 1 * COB C - cos a cos'] C - Cos'}a- cos' 1 + 1 4 sin f a sin } 6 cos C 2 sin a sin fb cosc cos 6 (6). ......

·AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 73 From this result we can deduce two other results, in the same manner as (4) and (5) were deduced from (3); or we may observe that the right-hand member of (6) can be obtained from the right-hand member of (3) by writing 7-a and #-6 for a and b respectively, and thus we may deduce the results more easily. We shall have then sin' (C-1)= cos 1 sin (8-a) sin 1 (8-6) cos (8-c) sin ja sin fb cosc sin 1 scos (8 a) cos } (8 — b) sin 1 (8-c) cos' (IC-LEsin a sinib cosc EXAMPLES. 1. 'Find the angles and sides of an equilateral triangle whose area is one-fourth of that of the sphere on which it is described. 2. Find the surface of an equilateral and equiangular spherical polygen of n sides, and determine the value of each of the angles when the surface cquals half the surface of the sphere. 7 3. If a -6 and c= shew that E=cos 9. T in 4. If the angle C of a spherical triangle be a right angle, shew that sin } E = sin a sin } 6 sec fc, cos} E = cos } a cos 16 sec 1 c. 5. If the angle C be a right angle, shew that sinc sinº sina a cos E = COS C cOS a COS O sin a 71 6. If a=l and C: shew that tan E= 2' 2 cosa 7. Tho sum of the angles in a right-angled triangle is less than four right angles. 8. Draw through a given point in the side of a spherical triangle an arc of a great circle cutting off a given part of the triangle.

·74 EXAMPLES. 9. In a spherical triangle if cos C=-tantan 2 C=A+B. om 6 2' then 10. If the angles of a spherical triangle bo together equal to four right angles cosº a + cos' } b + cos' lc =1. 11. If t, Te, Tg be the radii of three small circles of a sphere of radius r which touch one another at P, Q, R, and A, B, C be the angles of the spherical triungle formed by joining their centres, arca PQR=(A cos r, + B cosr, +C cos r,-)re. 12. Shew that sin s= sin 1 E sin (4 - 1 E) sin (B-1 E) sin (C-5 E) 2 sin | A sin 1 B sin C 13. Given two sides of a spherical triangle, determire when the area is a maximum. 14. Find the area of a regular polygott of a given number of sides formed by arcs of great circles on the surface of a sphere ; and hence deduce that, if a be the angular radius of a small circle, its area is to that of the whole surface of the sphere as versin å is to 2. 15. A, B, C are the angular points of a spherical trianglo; A', B", C" are the middle points of the respectively opposite sides. If E be the spherical excess of the triangle, shew that cos B'C cos CL cos 1 E= COB A'B' cos C costa cos 6 16. If one of the arcs of great circles which join the midlle points of the sides of a spherical triangle be a quadrant, shew that the other two are also quadrants.

·( 75 ) IX. ON CERTAIN APPROXIMATE FORMULÆ. 104. Wo shall now investigate certain approximatformulæ which are often useful in culculating spherical triangles when the raulius' of the sphere is large compared with the lengths of the sides of the triangles. 105. Given two sides and the included angle of a spherical, triangle, to find the angle between the chords of these sides. D E B cos EF Let AB, AC be the two sides of the triangle ABC; let O be the centre of the sphere. Describe a sphere round A as a centre, and suppose it to meet A0, AB, AC at D, E, F respectively. Then the anglo EDF is the inclination of the planes OAB, OAC, and is therefore equal to A. From the spherical triangle DEF = cos DE cos DF+ sin DE sin DF cos A; and DE=} (x-c), DF=} (-6); therefore cos EF = sin { b sin {c+cos 16 cos fc cos A. If the sides of the triangle are small compared with the radius of the sphere, EF will not differ much from A; suppose EF=A-0, then approximately cos EF = cos A + sin A ;

·76 ON CERTAIN APPROXIMATE FORMULÆ. and sin 1 sin fc= sin' (6+c) – sin': (6-c), cos 1 b cos 1 c= cos' 1 (b + c) – sin' (6-c); therefore cos A + 8 sin A = sin' 1 (6+ c) – sinº 1 (1 - c) + {1-sin" }(0+) – xin" }() –-)}ces COS A ; therefore sin A = (1 - cos A) sinº 2 (6 + c)-(1 + cos A) sino 1 (6-c), therefore 0 = tan 4 A sinº 1 (6+c) - cot! A sinº 1 (b -- c). a This gives the circular measure of 8; the number of seconds in the angle is found by dividir.; the circular incasure by the circular measure of one second, or approximately by the sine of one second (Plane Trigonometry, Art. 123). If the lengths of the arcs corresponding to a and 6 respevtively be a and B, and r the rulius of the sphere, we have and as the circular measures of a and / respectively; and the lengths of the sides of, the chordal triangle B are 2r sin and 2r sin respectively. Thus when the sides of 27 2r the spherical triangle and the radius of the sphere are known, we can calculate the angles and sides of the chordal triangle. 7 r a 4 106. Legendre's Theorem. If the sides of a spherical triangle be small compared with the radius of the sphere, then each angle of the spherical triangle exceeds by one third of the spherical esccess the corresponding angle of the plane triangle, the sides of which are of the same length as the arcs of the spherical triangle. Let A, B, C be the angles of the spherical triangle"; a, bet the sides ; r the radius of the sphere; , B, y the lengths of the β γ arcs which form the sides, so that are the circular measures of a, b, c respectively. Then q r ·ON CERTAIN APPROXIMATE FORMULÆ. 77 COS A = cos a COS 6 COS C sin b sin c now COS A -1- 2r? 247* a sin a=. . 1 623 Similar expressions hold for cost and sin b, and for cos c and sinc respectively. Hence, if we neglect powers of the circular measure above the fourth, we have 1 BY g* 1 2nd 2474 2r 2+r PY BP yok cos A 1 1 1 (1-P) (1-6) (Bº+y* -a") + umo(a* 239 = y* – 68*y?) (1-let) (a-80yº-6BYx}{1+ By pos 1 S 8² +8 Bi+y* g + 1 8 + 2By 1278 678 Be+ge - a* a* + Be + g * - 2a*ß* – 2B'ye – 2y®a 2By 24 Bypas Now let A', B', C' be the angles of the plane triangle whose sides are a, b, y respectively; then ß + ge - a 2By By sin'A' thus cos A 8 COS A' - = cos A'. Gr? Suppose A- A+; then cos A = cos A' - sin A' approximately; s therefore 6r² 3r By sin A ·78 ON CERTAIN APPROXIMATE FORMULÆ. where S denotes the area of the plane triangle whose sides are a, b, y. Similarly S S B=B' + and C=0" 378 زدن hence approximately S' S A + B + C = A' + B' + C'+ = T+ godt; S therefore is approximately equal to the spherical excess of the spherical triangle, and thus the theorem is established. It will be seen that in the above approximation the area of the spherical triangle is considered equal to the area of the plane triangle which can be fornied with sides of the same length. 107. Legendre's Theorem may be used for the approximate solution of spherical triangles in the following mamer. (1) Suppose the three sides of a spherical triangle known; then the values of a, b, y are known, and by the formula of Plane Trigonoinetry we can calculate S and A, B, C'; then 1, B, C are knowr. from the formula S A = A' + 32 B=B'+ S 3r- C=" + S 3. (2) Suppose two sides and the included angle of a spherical triangle known, for example A, 1, c. Then S =}By siń A' = 4 By sin A approximately. S Then A' is known from the formula A= A - Thas in the 378. plane triangle two sides and the included angle are known; therefore its remaining parts can be calculated, and then those of the spherical triangle become known.

·ON CERTAIN APPROXIMATE FORMULÆ. 79 ß (3) Suppose two sides and the angle opposite to one of them in a spolerical triangle known, for example A, a, b. Then B sin B sin A = sin A approximately; and C". 4--B'=#-A-B approximately; then $ = faß sin C'. Hence d' is known and the plane triangle can be solved, since 'two sites and the angle opposite to one of them are known. (4) Suppose two angles and the included side of a spherical triangle knowil, for example, 4, B, C. go sin A' sin B' y* sin A sin B Then S. nearly. sin ('+D") 2 sin (A + B) llence in the plane triangle two angles and the included side are kuowa. (5) Suppose two angles and the side opposite to one of them in a pherical triangle known, for example 4, B, a. Then ("'-- A' B'= 7-4-B, approximately, and a® sin B' sin (" S 2 sin (B+C')' which can be calculated, since B' and ' are approximately known. 108. The importanco of Legendre's Thcorem in the application of Spherical Trigonometry to the measurement of the Earth's surfiice has given rise to some developments of it which enable us to test the degree of exactness of the approximation. We shall finish the present Chapter with some of these developments, which will serve as exercises for the student. We have seen that ap- S proximately the spherical excess is equal to ge and we shall begjn with investigating a closer approximate formula for the spherical excess.

·80 ON CERTAIN APPROXIMATE FORMULA 109. To find an approximate value of the spherical excess. Let E denote the spherical excess; then sin ja sin fb sin C sin I E cos C therefore approximately ; . therefore sin E = sin c (1-7)(-) :-) =sinc> (1+%**) ***) (c +) )...(2). αβ E = sin c (1 ..(1), and sin C = sin(C" + JE =sin C'+ E cos C . = sin C'+ sin Cocos C" aß a + + ßa Esin C'11+ 3 From (1) and (2) a® + B + E = sin C αβ 2r 1 + 2473 Hence to this order of approximation the area of the spherical triangle exceeds that of the plane triangle by the fruction a* + Be + of the latter. 24708 sin A 110. To find an approximate value of sin B. sin a Sin A Sin B sin b ; a a (1 hence approximately sin A sin B $(1-6 br 120r* BP BY Og 12070* ·ON CERTAIN APPROXIMATE FORMULÆ. 81' BEO) 6 god a* BP 1 a'ß* B* B 120r Gr" 367 120p 3674 Be-al a-ßt, Bø (B8 - 2)) 1+ 678 120got 3674 . BE a a + BA - {1 = {1+ - {1+ (1+) :( B* 1+ 6r² 6r 207 )} 111. To express cot B-cot A approximately. 1 sin B Cot B-cot A (cos Bcos A); sin B sin A hence, approximately, by Art. 110, . 1 cot B-cot A (cos B. ß cos de Ba'-BP 67** cos 4). sin B a - y -as COS A Now we have shewn in Art. 106, that approximately ß8 + a* +84 + q* – 2c*ß* — 2B*y* – 2y*a* 2By 24Bype B a-BP therefore cos B approximately, ay a'- B B + y -a and cot B-cot A ay sin B 1278 al-B ß8 + goes 12708 COS A a - ay sin B ay sin B .112. The approximations in Arts. 109 and 110 are true so far as terms involving got ; that in Art. 111 is true so far as terms involving gre, and it will be seen that we are thus able to carry the approximations in the following Article so far as terms involving ra. T. S. T. G ·82 ON CERTAIN APPROXIMATE FORMULA 113. To find an approximate value of the error in the length of a side of a spherical triangle when calculated by Legendre's Theorem. Suppose the side ß known and the side a required; let 3u denote the spherical excess which is adopted. Therr the approximato B sin (4M) value is taken for the side of which a is the real sin ( BM) B sin (4 - Me) value. Let a a a ; wo have then to find a apsin ( BM) proximately. Now approximately sin A - u cos A- Mis sin A 2 sin (A -- M) sin (B-A) 2 sin B-M cos B - д 2 sin B sin A sin B M cot A - - Mucot B (1 { )(1-4 1+k (cot B – cot 4) + p? cot B (cot B-*cot 4)} sin A sin B sin A sin B ll sin A sin B (cot B-cot A) (1 + y cot B). Also the following formulæ aro true so far as terms involving your ; sin A sin BB $= (1+"). . a-BP cot B-cot A ay sin B (1 B8 + ge- 12go --7), 16M cot B=1+ . Hence, approximately, sin A al-B (cot B - oot A)(1 + p cot B) sin B By sin Bº ·ON CERTAIN APPROXIMATE FORMULÆ. 83 Therefore B sin A sin B hi (a' - B9) C = a y sin B 1 a (ße - a®) ou 6 3a' - 789) 60r 70*}, by by Art. 110. ay sin B 28 > we obtain ay sin B If we calculate pe from the formula 6r* a (B-a) (3a9-7B") 360r* If we calculate p from an equation.corresponding to (1) of Art. 109, we have 3B'-a- 1+ 6r² 2478 ay sin B a therefore (BP– a") (a* + ß* – 5y!). CE 720p* MISCELLANEOUS EXAMPLES. > 1. If the sides of a spherical triangle AB, AC be produced to B', C', so that BB', CC" are the semi-supplements of AB, AC respectively, shew that the arc B'C' will subtend an angle at the centre of the sphere equal to the angle between the chords of AB and AC. 2 9 2. Deduce Legendre's Theorem from the formula A sin } (a +6 - c) sin 1 (c + a -- 6) tan 2 sin (6 + c-a) sin } (a + b + c) * 3. Four points A, B, C, D on the surface of a sphere are joined by arcs of great circles, and E, F are the middle points of the arcs•AC, BD: show that Cos AB+cos BC +cos CD + cos DA = 4 cos AE cos BF cos FE. 4. If a quadrilateral ABCD be inscribed in a small circle on a sphere so that two opposite angles A and C may be at opposite G 2 ·84 MISCELLANEOUS EXAMPLES. extremities of a diameter, the sum of the cosines of the sides is constant. 5. In a spherical triangle if A =B= 2C, shew that cos a cos cos = cos (c+) . 6. ABC is a spherical triangle each of whose sides is a quadrant; P is any point within the triangle: shew that cos PA COS PB cos PC + cot BPC cot CPA cot APB -0, and tan ABP tan BCP tan CAP=1. 7. If O be the middle point of an equilateral triangle ABC, and P any point on the surface of the sphere, then (tan PO tan 0A) (cos PA +cos PB + cos PC) cos'PA + cos'PB+cos PC-cos P Acos PB-cosPBcos PC-cos PCcos P'A 8. If ABC be a triangle having each side a quadrant, О the pole of the inscribed circle, P any point on the sphere, then (cos PA + cos PB + cos PC) = 3 cos® PO. 9. From each of three points on the surface of a sphere arcs are drawn on the surface to three other points situated on a great circle of the sphere, and their cosines are a, b, c; a',', d'; a", 6", ". Shew that ab"C' + 'a'bc" + a"'c = ab'c" + a'b"C + abc'. 10. From Arts. 110 and 111, shew that approximately s log B = log a + log sin B - log sin A + (cot A-cot B). Зr 11. By continuing the approximation in Art. 106 80 as to include the terms involving r, shew that approximately By sin A' Byla' – 3B – 3y) sin'A' cos A = cos 6r 180r4 COSA 12. From the preceding result show that if A = A' + then approximately By sin A A = 7B* + 7y + 1+ Gr 120 +4).

·( 85 ) X. GEODETICAL OPERATIONS. 114. One of the most important applications of Trigonometry, both Plane and Spherical, is to the determination of the figure and dimensions of the Earth itself, and of any portion of its surface. We shall give a brief outline of the subject, and for further information refer to Woodhouse's Trigonometry, to the article Geodesy in the English Cyclopædia, and to Airy's treatise on the figure of the Earth in the Encyclopædia Metropolitana. For practical knowledge of the details of the operations it will be necessary to study some of the published accounts of the great surveys which have been effected, in different parts of the world, as for example, the Account of the measurement of two sections of the Meridional arc of India, by Lieut-Colonel Everest, 1847; or the Account of the Obserdations and Calculations of the Principal Triangulations in the Ordnance Survey of Great Britain and Ireland, 1858. 115. An important part of any survey consists in the measurement of a horizontal line, which is called a båse. A level plain of a few miles in length is selected and a line is measured on it with every precaution to ensure accuracy. Rods of deal, and of metal, hollow tubes of glass, and steel chains, have been used in different surveys; the temperature is carefully observed during the operations, and allowance is made for the varying lengths of the rods or chains, which arise from variations in the temperature. 116. At various points of the country suitable stations are selected and signals erected; then by supposing lines to be drawn connecting the signals, the country is divided into a series of triangles. The angles of these triangles are observed, that is, the angles which any two signals subtend at a third. For example, suppose A and B to denote the extremities of the base, and C a ·86 GEODETICAL OPERATIONS. signal at a third point visible from A and B; then in the triangle ABC the angles ABC and BAC are observed, and then AC and BC can be calculatod. Again, let D be a signal at a fourth point, such that it is visible from C and A ; then the angles ACD and CAD are observed, and as AC is known, CD. and AD can be calculated 117. Besides the original base other lines are measured in convenient parts of the country surveyed, and their measured lengths are compared with their lengths obtained by calculation through a series of triangles from the original base. The degree of closeness with which the measured length agrees with the calculated length is a test of the accuracy of the survey. During the progress of the Ordnance Survey of Great Britain and Ireland, several lines have been measured; the last two are, one near Lough Foyle in Ireland, which was measured in 1827 and 1828, and one on Salisbury Plain, which was measured in 1819. The line near Lough Foyle is nearly 8 miles long, and the line osa Salisbury Plain is nearly 7 miles long; and the difference between the length of the line on Salisbury Plain as measured and as calculated from the Lough Foyle base is less than 5 inches (An Account of the Observations ...page 419). 118. There are different methods of effecting the calculations for determining the lengths of the sides of all the triangles in the Burvey. One method is to use the exact formulæ of Spherical Trigonometry. The radius of the Earth may be considered known very approximately; let this radius be denoted by r, then if a be the length of any are the circular measure of the angle. which the arc subtends at the centre of the earth is The formule of Spherical Trigonometry give expressions for the trigonometrical funotions of so that may be found and then a. Since in a practico, is always very small, it becomes necessary to pay ·GEODETICAL OPERATIONS. 87 attention to the methods of securing accuracy in calculations which involve the logarithmic trigonometrical functions of small angles (Plane 1'rigonometry, Art. 205). Instead of the exact calculation of the triangles by Spherical Trigonometry, various methods of approximation have been proposed ; only two of these methods however have been much used. One method of approximation consists in deducing from the angles of the spherical triangles the angles of the chordgl triangles, and then computing the latter triangles by Plane Trigonometry (see Art. 105). The other method of approxination consists in the use of Legendre's Theorem (see Art. 106). 119. The thrce methods which we have indicated were all used by Dolambre in calculating the triangles in the French survey (Base du Système détrique, Tome III. page 7). In the earlier operations of the Trigonometrical survey of Great Britain. and Ireland, the triangles were calculated by the chord method; but this has been for many. years discontinued, and in place of it Legenchre's Theorem has been universally adopted (An Account of the Observations page 244). The triangles in the Indian Survey are stated by Lieut-Colonel Everest to be computed on Legendre's Theorem. (An Account of the Measurement ... page cLvII.) 120. If the three angles of a plane triangle be observed, the fact that their sum vught to be equal to two right angles affords a test of the accuracy with which the observations are made. We shall proceed to show how a test of the accuracy of observations of the angles of a spherical triangle formed on the Earth's surface may be obtained by means of the spherical excess. 121. The area of a spherical triangle formed on the Earth's surface being known in square feet, it is required to establish a rule for computing the spherical excess in seconds. Lot ne be the number of seconds in the spherical excess, 8 the number of square feet in the area of the triangle, the number of ·88 GEODETICAL OPERATIONS. foot in the radius of the Earth.' Then if E be the circular mean sure of the spherical excess, 8 = Er", na n and E = 180.60.60 206265 approximately; therefore nr 206265 Now by actual measurement the mean length of a degree on the Earth's surface is found to be 365155 feet; thus TT 180 365155. With the value of r obtained from this equation it is found by logarithmic calculation, that log n = log 8 - - 9.326774. Hence n is known when : is known. This formula is called General Roy's rule, as it was used by him in the Trigonometrical survey of Great Britain and Ireland. Mr Davies, however, claims it for Mr Dalby. (See Hutton's Course of Mathematics, by Davies, Vol. II. p. 47.) 122. In order to apply General Roy's rule, we must know the area of the spherical triangle. Now the area is not known exactly unless the elements of the spherical triangle are known exactly; but it is found that in such cases as occur in practice an approximate value of the area is sufficient. Suppose, for example, that we use the area of the plane triangle considered in Legendre's Theorem, instead of the area of the Spherical Triangle itself; then it appears from Art. 109, that the error is approximately a® + Be + x denoted by the fraction of the former area, and this 24.gode fraction, is less than .0001, if the sides do not exceed 100 miles in lengtih. Or again, suppose we want to estimate the influence of errors in the angles on the calculation of the area; let the ·GEODETICAL OPERATIONS. 89 aß sinc circular measure of an error be h, so that instead of 2 aß sin (C + h) we ought to use ; the error then bears to the area 2 approximately the ratio expressed by h cot C. Now in modern observations h will not exceed the circular measure uf a few seconds, so that, if C be not very small, h cot C is practically insensible, > 123. The following example was selected by Woodhouse from the triangles of the English survey, and has been adopted by other writers. The observed angles of a triangle being respectively 42°. 2. 32", 67º. 55'. 39", 70°. 1'. 48", the sum of the errors made in the observations is required, supposing the side opposite to the ingle A to be 27404•2 feet. The area is calculated from the exas sin B sinc pression, and by General Roy's rule it is found 2 sin A •23.. Now the sum of the obsexved angles is 180°-1", and as it ought to have been 180° + •23", it follows that the sum of the errors of the observations is 1"-23. This total error may be distributed among the observed angles in such proportion as the opinion of the observer may suggest ; one way is to increase egch of the observed angles by one-third of 1"-23, and take the angles thus corrected for the true angles. that n= 124. An investigation has been made with respect to the form of a triangle, in which errors in the observations of the angles will exercise the least influence on the lengths of the sides, and although tho reasoning is allowed to be vague it may be deserving of the attention of the student. Suppose the three angles of a triangle observed, and one side, as a, known, it is required to find the form of the triangle in order that the other sides may be least affected by errors in the observations. The spherical excess of the triangle may be supposed known with sufficient accuracy for practice, and if the sum of the observed angles does not exceed two right angles by the proper spherical excess, let these angles be altered by adding the same quantity to ·90 GEODETICAL OPERATIONS. each, so as to make their sum correct. Let A, B, C be the angles thus furnished by observation and altered if necessary; and let SA, 8B and 8C denote the respective errors of A, B and C. Then 84 +8B+8C = 0, because by supposition the sum of A, B and C is correct. Considering the triangle as approximately plane, the true value of the side c is a sin (C+8C) a sin (C+8C) that is, sin (4 +8A) sin (A - SB - 00) Now approximately sin (C+8C)=sin C + 8C cos C, (Plane Trig. Chap. XI.), sin (4 - SB-8C) = sin A - (8B + 8C) cos A.. a sin C sin A Hence approximately + 8C cotc}{1-(8B+ 8C) cot 1 ot a)" 1 +8B cot A + 8C (cot C + cot 4) }; . sin (A + C) sin B sin A sin Ćsin A sin C approximately. a sin C sin A and cot C + cot A : Hence the error of c is approximately a sin B sin A 8C + a sin C cos A SB. sin' Similarly the error of b is approximately a sin C sinA SB + a sin B cos A sin' A 8C. Now it is impossible to assign exactly the signs and magnitudes of the errors 8B and 8C, so that the reasoning must be vague. . It is obvious that to make the error small sin A must not be nati: And as the sum of 81, 8B and 8C is zero, two of them must have the same sign, and the third the opposite sign; we may therefore consider that it is more probable that any two as 8B and 8C have different signs, than that they have the same sign.

·GEODETICAL OPERATIONS. 91 If 8B and 8C have different signs the errors of b and c will be less when cos A is positive than when cos A is negative ; A therefore ought to be less than a right angle. And if 8B and 8C are probably not very different, B and C should be nearly equal. These conditions will be satisfied by a triangle differing not much from an equilateral triangle. If two angles only, A and B, be observed, we obtain the same expressions as before for the errors in and c; but we have no reason for considering that 8B and 8C are of different signs rather than of the same sign. In this case then the supposition that A is a right angle will probably make the errors smallest. 125. The preceding article is taken from the Treatise on Trigonometry in the Encyclopædia Metropolitana. The least satisfactory part is that in which it is considered that 8B and 8C may be supposed nearly equal; for siņce 84 +8B + 8C = 0, if we suppose 8B and 8C nearly equal and of opposite signs, we do in effect si ppose 8A = 0 nearly; thus in observing three angles, we suppose that in one observation a certain error is made, in a second observation the same numerical error is made but with Pat opposite sign, and in the remaining observation no error is made. 126. We have hitherto proceeded on the supposition that the Earth is a sphere ; it is however approximately a spheroid of small eccentricity. For the small corrections which must in consequence be introduced into the calculations we must refer to the works named in Art. 114. One of the results obtained is that the error caused by regarding the Earth as a sphere instead of a spheroid increases with the departure of the triangle from the well-conditioned or equilateral form (An Account of the Observations ... page 243). Under certain circumstances the spherical excess is the same on a spheril as on a sphere (Figure of the Earth in the Encyclopaedia Metropolitana, pages 198 and 215). 127. In geodetical operations it is sometimes required to determine the horizontal angle between two points, which are at a

·92
GEODETICAL OPERATIONS.
small angular distance from the horizon, the angle which the
objects subtend being known, and also the angles of elevation
or depression.
B
D
Suppose 04 and OB the directions in which the two points
are seen from 0; and let the angle AGB be observed. Let OZ be
the direction at right angles to the observer's horizon ; describe
a sphere round 0 as a centre, and let vertical planes through 04
and OB meet the horizon at OC and OD respectively : then the
angle COD is required.
Let AOB = 0, COD = 0 + X, AOC = h, BOD=k; from the
triangle AZB
cos 0 - cos ZA cos ZB
COB AZBE
sin ZA sin ZB
cos 6-sin h sin k
cos h cos k
;
and cos AZB = cos CO.D=cos (+ x); thus
cos (8 + x) =
cos 0 -sin h sin k
cos h cos
This formula is exact; by approximation we obtain
cos 0 - 2 sin 0
Cos 0 - hk
1-(k+ kyi

·ON SMÅLL VARIATIONS. 93 therefore x sin 0 =hk -- }(h* + k*) cos 0, nearly, 2hk - (h® + k*) (cos' 10-sin' 10) 2 sin 0 and -(h+k)* tan 0-1 (h-k) cot 0. This process, by which we find the angle COD fronı the arrgle AOB, is called reducing an angle to the horizon. XI. ON SMALL VARIATIONS IN THE PARTS OF A SPHERICAL TRIANGLE. 128. It is sometimes important to know what amount of error will be introduced into one of the calculated parts of a triangle by reason of any small error which may exist in the given parts . We will here consider an example, 129. A side and the opposite angle of a spherical triangle remain constant : determine the connexion between the small variations of any other pair of elements. Suppose C and c to remain constant. (1) Required the connexion between the small variations of the other sides. We suppose a and 6 to denote the sides of one triangle which can be formed with C and c as fixed elements, and a + da and 6+86 to denote the sides of another such triangle; then we require the ratio of da to 86 when both are extremely small. We have and cos c= cos a cos 6+ sin a sin 6 cos C, COS C= cos (a + da) cos (6 +86) + sin (a + da) sin (6.+86) cos C; cos (a + da)= cosa - sin a da, nearly, sin (a + da)= sin a + cos a da, nearly, also and ·94 ON SMALL VARIATIONS IN THE with similar formulæ for cos (6 +86) and sin (6 +85). (Soe Plane Trigonometry, Chap. xn.) Thus cos c= (cos a – sin a da) (cos 6 - sin 686) +(sin a + cos a da) (sin 6 + cos 686) cos C. Hence by subtraction, if we neglect the product da 86, 0 = 8a (sin a cos 6 - cos à sin b cos C) +86 (sin b cos a cos b sin a cos C); this gives the ratio of da to 86 in terms of a, b, C. We may express the ratio more simply in terms of A and B; for, dividing by sin a sin b, we get from Art. 44, da 86 cot B sin C + cot A sin C=0; sind sin a therefore da cos B + 60 cos A = 0. (2) Required the connexion between the small wariations of the other angles. In this case we may by means of the polar triangle deduce from the result just found, that 8A cos 6 + 8B cos,a=0); this may also be found independently as before, (3) Required the connexion between the small variations of a side and the opposite angle (A, a). Here sin A sin c = sin C sin a, and sin (4 + 8A) sin c=sin C'sin (a + da); hence by subtraction cos A sin c 8A = sin C cos a da, and therefore 84 cotA=&qcot a. (4) Required the connexion between the small variations of a side and the adjacent angle (a, B).

·PARTS OF A 'SPHERICAL TRIANGLE. go We have cot C sin B=cot c sin a cos B cos a.; proceeding as before we obtain cot C cos B 8B = cot c cos a da + cos B sin a da + cos a sin B&B; therefore (cot C cos B - cos a sin B) 8B = (cotc cos a + oos B sin a) Sa ; therefore GOS A cos 6 SB sa; sin C sin c therefore SB cos A da cot b sin B, 130. Some more examples are proposed for solution at the end of this Chapter; as they involve no difficulty they are left for the exercise of the student, EXAMPLES. sin c 1. In a spherical triangle, if C and c remain constant while a and b receive the small increments da and 86 respectively, shew that δα 86 sin c 0 where n= J(1 - nsinº a) J(1 - nº sinº 6) 2. If C and c remain constant, and a small change be made in a, find the consequent changes in the other parts of the triangle. Find also the change in the area. 3. Supposing A and c to remain constant, prove the following equations, connecting the small variations of pairs of the other elements : sin C 86 = sin a 8B, 86 sin c da tan C=8B sin a, da tan C=- 8C tan a, 86 cos C = da, 8B cos a = -8C.. 4. Supposing 6 and c to remain constant, prove the following equations connecting the small variations of pairs of the other elements : 8B tan C=BC tan B, da oot C =- 80 = 84 sin c sin B, 8A sin B cos C=-&B sin A, - 8C tan a, 8B sin ang ·96 EXAMPLES 5. Supposing B and C to remain constant, prove the following equations connecting the small variations of pairs of the other elements : 80 tan c= ốc tan b, SA cot c= 86 sin A, 8A = da sin b sin C, da sin B Cos C= 86 sin A. 6. If A and C are constant, and I be increased by a small quantity, shew that a will be increased or diminished according as c is less, or greater than a quadrant. IN XII. ON THE CONNEXION OF FORMULÆ PLANE AND SPHERICAL TRIGONOMETRY. 131. The student must bave perceived that many of the results obtained in Spherical Trigonometry resemble others with which he is familiar in Plane Trigonometry. We shall now pay some attention to this resemblance. We shall first shew how we may deduce formulæ in Plane Trigonometry from formulæ in Spherical Trigonometry; and we shall then investigate some theorems in Spherical Trigonometry which are interesting principally on account of their connexion with known results in Plane Geometry and Trigonometry. 132. Fron any formula in Spherical Trigonometry involving the elements of a triangle, one of them being a side, it is required to deduce the corresponding formula in Plane Trigonometry. Let , B, y be the lengths of the sides of the triangle, the radius of the sphere, so that , B, 1 are the circular measures α β γ of the sides of the triangle; expand the functions of re gine 70 a ß a 7 which occur in any proposed formula in powers of go respectively; then if we suppose q to become indefinitely great, ·CONNEXION OF FORMULE IN TRIGONOMETRY. 97 COS a COS A: the limiting form of the proposed formula will be a relation in Plane Trigonometry. For example, in Art. 106, from the formula cos 6 cos c sin b sin c we deduce BP + gye - a a* +34 +y* - 2a*ß* - 2B'y – 2y'a 2βγ 24 Byr" now suppose r to become infinite; then ultimately Bi+ge-a 2By and this is the expression for the cosine of the angle of a plane triangle in terms of the sides. Again, in Art. 110, from the formula cos 4 COS A ; sin a sin A in B 0 sin b a sin A a (Bi-a) we deduce sip BB 6pr8 now suppose r to become infinite; then ultimately sin A sin B B that is, in a plane triangle the sides are as the sines of the opposite angles. a > 133. To find the equation to a small circle of the sphere. The student can easily draw the required diagram. Let O be the pole of a small circle, S a fixed point on the sphere, *SX a fixed great circle of the sphere. Let OS=a, OSX=B; then the position of O is determined by meins of these angular co-ordinates a and B. Let P be any point on the circumference of the small circle, PS= 6, PSX = 6, 80 that 0 and $ are T. S. T. H ·98 CONNEXION OF FORMULE the angular co-ordinates of P. Let OP=r. Tuon from the triangle OSP cos r= cos a cos 0 + sin a sin 0 cos ($-B)............ (1), this gives a relation between the angular co-ordinates of any point on the circumference of the circle. . ! If the circle be a great circle then 'n ; thus the equation becomes 0= = cos a cos 0 + sin a sin 6 cos ($-B)......... (2). It will be observed that the angular co-ordinates liere uscd are analogous to the latitude and longitude which serve to deterinine the positions of places on the Earth's surface; O is the complement of the latitude and is the longitude. 1 134. Equation (1) of the preceding Article may be written 0 6 thus : cos 2 0 0 = cos a cos sin to 2 sin a sin COB ($-B). 2 9 ) COS Divide lay cos A 2 and rearrange; hence 9 tan o A 0 (cos " + cps a) – 2 tan sin a cos ($-B) + cos r - cos a == A Let tan and tan denote the values of tan found from 2 this quadratic equation ; then by Algebra, Chapter XXII. . O, atr tan tan tan tan 2 2. COS r + cos a 2 2 CORT COS a a 1 - 074 Thus the value of the product tan tan is independent of ; 2 this result corresponds to the well-known property of a circle in Plane Geometry which is demonstrated in Euclid 111. 36 Corillding. 135. Let three arcs 0A, OB, OC meet at a point. From any point P in OB draw PM perpendicular to 04, and PN perpen- 1 dicularuto OC. The student can easily draw the required diagram.

·IN PLANE AND SPHERICAL TRIGONOMETRY, 99 Then, by Art. 65, sin PM = sin OP sin AOB, sin PN = sin OP sin COB; sin PN sin AOB therefore sin PN sin COB Thus the ratio of sin PM to sin PN is independent of the position of P on the arc OB. 136. Conversely suppose that from any other point p arcs pm and pn are drawn perpendicular to 04 and OC respectively; then if sin PM sin pm sin pre sin PN' it will follow that p is on the same great circle as O and P. 1 and sin x= 2 sin xy + sin a 137. From two points P, and P, arcs are drawn perpendicular to a fixed aro; and from a point P on the same great circle as P, and P, a perpendicular is drawn to the same fixed arc. Let PP, = 0, and PP,= 0,; and let the perpendiculars drawn from P, Pin and P, be denoted by x, ** Then will sin sin , sin (0, +0,2) sin («, +0) Let the arc P, Pg, produced if necessary, cut tho fixed arc at a point 0; let a denote the angle between the arcs. We will suppose that P, is between 0 and P2, and that P is between P, and Pg. Then, by Art. 65, sin a sin OP, = sin a sin (OP-0,) sin a (sin OP cos 6, -cos OP sin 0,); sin a sin OP, = sin a sin (OP+0) sin a (sin OP cos 0, +cos OP sin 6.). Maltiply, the former by sin 0g, and the lattor by sin ,, and add ; thus sin 0, sin x, + sin 0, sin a, = sin(@, +0,) sin a sin OP sin (0, +0.) sin .. H? sina, sin og ·100 CONNEXION OF FORMULA The student should convince himself by examination that the result holds for all relative positions of P, P, and Pr, when due regard is paid to algebraical signs. 138. The principal use of Art. 137 is to determine whether three given points are on the same great circle; an illustration will be given in Art. 146. 139. The arcs drawn from the angles of a spherical triangle perperuicular to the opposite sides respectively meet at a point. E FA . sinn Let CF be perpendicular to AB. From F suppose arcs drawn perpendicular to CB and CA respectively; denote the former.by and the latter by n. Then, by Art. 135, sin sin FCB sin FCA But, by Art. 65, cos B= = cos CF sin FCB, cos A = cos CF sin FCA; sin & therefore cos B cos C COS A cos A cos C cos B sinn And if from any point in CF arcs are drawn perpendicular to CB and CA respectively, the ratio of the sine of the former perpensin dicular to the sine of the latter perpendicular is equal to by Art. 135. sinn ·IN PLANE AND SPHERICAL TRIGONOMETRY. 101 In like manner suppose AD perpendicular to BC; then if from any point in AD arcs are drawn perpendicular to-AC and AB respectively, the ratio of the sine of the former perpendicular to cos A cos C the sine of the latter perpendicular is equal to cos A cos B Let CF and AD met at P, and from P let perpendiculars be drawn on the sides a, b, c of the triangle; and denote these perpendiculars by x, y, z respectively : then we have shewn that sin a sin y cos B cos C COS A COS sin y and that cos A cos C cos A cos Bi sin hence it follows that sin 2 cos B cos cos B cos A sin and this°shows that the poiat P is on the arc drawn from B perpendicular to AC. Thus the three perpendiculars meet at a point, and this point is determined by the relations sin a sin y sin z cos B cos o cos C cos A cos A cos B 140. In the same manner it may be shewn that the arcs drawn from the angles of a spherical triangle to the middle points of the opposite sides meet at a point; and if from this point arcs %, y, z are drawn perpendicular to the sides a, b, c respectively, sin a sin y sin % sin B sin sin C sin A sin A sin B 141. It is known in Plane Geometry that a certain circle touches the inscribed and escribed circles of any triangle; this circle is called the Nine points circle: see Appendix to Euclid, pages 317, 318, and Plane Trigonometry, Chapter XXIV.

·102 CONNEXION OF FORMULA We shall now show that a small circle can always be determined on the sphere to touch the inscribed and eścribed circles of any spherical triangle. 142. Let a denote the distance from A of the pole of the small circle inscribed within a spherical triangle ABC. Suppose that a small circle of angular raulius p touches this inscribed circle internally; let B be the distance from 4 of the pole of this touching circle ;, let y be the angle between arcs drawn from A to the pole of the inscribed circle and the pole of the touching circle respectively. Then we must have cos (p-r) = cos a cosB + sin a sin ß cos y ...(1). Suppose that this touching circle also touches externally the escribed circle of angular ralius r,; then if a, denote tho distance from A of the pole of this escribed circle, we must have cos (o + r.) = cpsa, cos B + sin a, ; (2). Similarly, if a, and q, denote the distances from A of the poles of the other escribed circles, in order that thic touching circle may touch these escribed circles externally, we must also have cos (o + r) Cosa, cos ß + sin ag sin ß cos (5.-Y.........(3), Cos (0+r) = cos a, cos B + sin a, sin B cos (5 + y...... (4). We shallöshew that real values of p, B, and y can be found to satisfy these four equations. sin ß cos y Eliminate cos y from (1) and (2); thus cos p (cos r sin a, cos r, sin a) + sin p (sin r sin a, + sin r, sin a) = cosB (cos a sin a, - cosa, sin a)......(). Suppose that the inscribed circle touches AB at the distance mi from A, and that the escribed circle of angular radius r, touches AB at the distance m. from A. Then, by Art. 65, ·IN PLANE AND SPHERICAL TRIGONOMETRY. 103 A cot a=cot m cos cos a = cos r cos m, sin r=sin a sin A 2 COS 1 cota 1 A therefore COS . sul a cOS m sin m a and with m,. Similarly we may connect cbtain from (5) Thus we А 1 1 COS P COS 1/16 + 2 sin o sin A 2 sin m sin m m, A = cos ß cos (cot m - cot m.); 2 therefore cosp (sin m, sin n) + 2 sin p sin m sin m, tan 2 = cos ß sin (m, - m). =S-a, and mi =8; there- But by Arts. 89 and 90 we have m foro by the aid of Art. 45 we obtain a b + c • 2 cos p sin COS + 2n sin p=cos B sin a ............(6), mg cot where n has the meaning assigned in Art. 46. In like manner if we diminate sin y between (3) and (4), partting mo for 8-C, and та for s-b, we obtain A sin cosp (sin m, + sin mg) – 2 sin p sin ma = cos ß sin (me + m), 6 therefore 2 cosp sin 2n sinp=cosß sina ..(7). 2 2 From (6) and (7) we get 7 sin sin 2 2 1 tan R, by Art. 92 .........(8), · 2 с COS a sin tan p n2 ܤ ܗܘ;ܘܗ COS 6 2 COS cos p 2 and cos B (9). COS 2 ·104 CONNEXION OF FORMULA or COS 80 6 We may suppose that cos is not less than cos 2 2 that we are sure of a possible value of cos B from (9). It remains to show that when p and ß are thus determined, all the four fundamental equations are satisfied. It will b3 observed that, p and B being considered known, cos y can be found from (1) or (2), and sin y can bo found from (3) or (4): wu must therefore shew that (1) and (2) give the same value for cos y, and that (3) and (4) giro tho same value for sin y; and we must also shew that these values satisfy tlo condition cos' y+siny=1. From (1) we have cos p sin r cos B cot r + tan p- cos m cotr sin fcos y, sin a COS р that is, COS р sin А 2 6 cos (8 - a) sin 8 cos COS 2 sin 8 + sinf a sin sin dc n cos COS P COS a COS n a = sin B cos'y i this reduces to A 6 sin sin (b + c) cos b + c 2 sin sin ß cos y : 2 cos 2 and it will be found that (2) reduces to the same; so that (1) and (2) give the same value for cos y. In like manger it will be found that (3) and (4) agree in reducing to A 6 sin (0-6) cos P 6 2 = sin ß sin y cos sin 2 cos 2 eos COS n ·IN PLANE AND SPHERICAL TRIGONOMETRY. 105 It only remains to shew that the condition cos'y +sin' y=1 is satisfied. cos ß Put k for COS that is for 2 c COS 2 cOS P a COS put X for cot r{1 - k cos (8-a)}, and Y for cot r, {1 – k cos 8}. Then (1) and (2) may be written respectively thus: A 2 8γ sin ßcos y (11). ........... (Y cosp - sin p) sin From (10) and (11) by addition A (X+Y) sin cosp=2 sin ß cos y; A + sinº But from (10) and (11) dy subtraction (X - Y) cosp=-2 sin p; therefore (.1* + Y) cos'p 4 sinºp+ 2XY cos p. Substitute in (12) and we obtain A sinº8 cos® y = (sinºp+XY cos® p) sii. ...... (13). Again, put X, for cotr, {1 - k cos (8-c)}, and Y, for cotr, {1 - k cos (8-6)}. Then (3) and (4) may be written respectively thus : A (X, cos p-sin p) cos = sin ß sin y (14), 2 sin B sing (15). ....... (Y, cos p-sin p) cos 2 From (14) and (15) by subtraction (4, -Y.) cos ܬ ܠܐܟܬ cos p = 2 sin ß sin y, ·106 CONNEXION OF FORMULÆ and from (14) and (15) by addition, (X, + Y) cos p=2 sin p, whence sin'B sin'y=(sinºp-X,Y, cosºp)cos® (16). ....... (xYsin 1.4) $ 1 – K*- XY sinº - 1,1, cose Hence from (13) and (16) it follows that we have to establish the relation sin' B =sin' p +(XY sinº --I, Y, cosa cosa 2 p. But sin ß=1- cos® B -- sin' p + cosp-k* cose p, so that the relation reduces to А А 2' Now TY sin’ A cotrcotr, {1- kcoss}{1 - k cos(8 - 0) sin(x-)sin (x-r) 2 sin b sinc {1 - k cos 8} {1 – k cos (8 - a)} sin 6 sinc A cos 2 sin lusin c Subtract the latter from the former; then we obtain {cos (8-6) + cos (8 – c) - COS 8 – cos (8 - a)} sin b sin c {cos 8 cos (8-a) - cos (8 – 6) cos (8-c)}, sin b sinc a 2k cos C that is 2 sin o sin 6 COS 6+ COS o to 2 K* btcta COS 2 5+C-a 2 a + C-6. COS COS 2 a + 6-c) 2 slu 6 sinccos COS 2 C + sin:c-6 b b 4 sin sin cos 2 K* that is sin b sin c sin o sin cl that is 1 -; which was to be shewn. sin 2 ·a cos COS COS P COS COS COS P ܘܗܘ COS COS cos IN PLANE AND SPHERICAL TRIGONOMETRY. 107 143. Thus the existence of a circle which touches the inscribed and escribed circles of any spherical triangle has been established. The distance of the pole of this touching circle from the angles B and C of the triangle will of course be determined by formulæ corresponding to (9); and thus it follows that 6 2 2 2 and 6 must both be less than unity. 141. Since the circle which has been determined touches the inscribed circle internally and touches the escribed circles externally, it is obvious that it must meet all the sides of the spherical triangle. We will now determine the position of the. points of nieeting. Suppose the touching sircle intersects the side AB at points distant and u respectively from A. Then by Art. 134 we have 6 λ M cos B 2 2 tan .. (1). 2 cosp + cos B 6 2 2 In the same way we must have by symmetry 6 λ 2 tan tan (2). 2 6 2 2 2 λ From (2), when we substitute the value of tan tan given 71),ewe obtain cos cos + cos – sinº λ M 2 2 2 tan + tan 2 6 6 si 2 2 2 a COS COS COS COSP tan a COS + cOS COS 2 COS COS COS C-M - 7 ...... a COS + COS COS a 25 6 2 2 C a COS (cos +COS COS 3) ·108 CONNEXION OF FORMULE 810 COS sin 2 COS COS COS 2 2 b COS sin 2 (3). ........ a COS + COS COB 2 a From (1) and (3) we sce that we may put 3 COS λ 2 2 tan 6 sin 2 2 COS cos = 2 (4). COS tan 6 COS sin 2 2 a 6 COS + cos COS 2 (5). .. 2 eine Similar formulæ of course hold for the points of intersection of the touching circle with the other sides. 145, Let , denote the perpendicular from the pole of tho touching circle on AB; then sin a= sin B sin (6 +y) = sin B (sin cos y + cos sin y) But from (2) and (3) of Art. 142 we have cospsin î sin B cos y= a 2-sin - sin 6 2 sin n , where Z=sin (e –a) – cos e sin (e – a) cos cos sec 8 A cos P COS 2 and sin B sin y= (2 , -sin sinsin), n c where 2 = sin (8-6) - cos (8-c) sin (8-6) cos 6 2 COS sec 1 ·IN PLANE AND SPHERICAL TRIGONOMETRY. 109 Therefore A 2 sin ==C6 ? {2 min® 4+2, cos" 4 - Bing win ain). Now Zin sin (8-a) sin (8-8) sin (8-c) =){1- and 2, cos? sin s sin (o 7 al sin (–B) {1- 5 COS 8 COS COS sin b sinc sec a} cos (8 – c) cos 6 2 COS sec a 2) Therefore Zsin? А 2 +2, cosa A 2 is equal to the product of sin (8-a) sin (8- • sin b sin c into 5) a sin (* --c)+sin 8 COS COS C sec 2. 2 10 (6-4) 688 +000 (6— c) sino? - cos coses sec. sin (28- (26–c)} - a + 6 C 6 sin (8 - «) sin (8-5) sin b sinc { sin COS 2 sin (: - a) sin (8-6) atb 5}{26 sin sin (a + b) cos sec 2 sin b sin . a+b a+b 6 sin (8-a) sin (8~5) sin COS COS 2 2 1 a sin b sin COS 2 a+b b a fb sin (8-a) sin(8--6) sinº sin sin (8-a) sin(8-5) sin? 2 2 a sin b sin COS 2 cocos sin 2 ·110 CONNEXION OF FORMULÆ Therefore a+b 2 sin sin (8-a) sin (8-0) a COSL sinsin, sin 6 2 sin = . 1 n 2 sin sin a sin 6 2 COSP a 5 sin sin sin C B 2 { 2 cos 3}; by (2) of Art. 54. 2 20 2 COSP Thus sin = sin sin sin cos (14 – B) sin p cos (A - B). Similar expressions hold for the perpendiculars from the pole of the touching circle on the other sides of the spherical triangle, 146. Let P denote the point determined in Art. 139; G the 'point determined in Art. 140, and V the pole of the touching circle. We shall now phew that P, G, and V are on a great circle. Let x, y, z denote the perpendiculars from N on the sides a, b, c respectively of the spherical triangle, let x, y, denote the perpendiculars from P; and be, Y, 2, the perpendiculars from G. Then by Arts. 145, 139, and 140 we have sina siny ginz cos (B-C) cos (C-A) cos (C-1) cos (A - B)' sin y, sin , cos B cos C cos C cos A cos A cos B' sin , sin leg sin %. siny, sin C'sin A sin B sin C sin A sin B Hence it follows that sin x=t, sin x, + t, sin X, sin y=t, sin y, + t sin ), sin 2 = +, sin 2, +t, sin 2, ·, IN PLANE AND SPHERICAL TRIGONOMETRY. 111 where and t, are certain quantities the values of which are not required for our purpose. Therefore by Art. 137 a certain point in the same great circle as P and G is at the perpendicular distances x, y, z from the sides a, b, c respectively of the spherical triangle : and hence this point must be the point N. 147. The resemblance of the results which have been obtained to those which are known respecting the Nine points circle in Plane Geometry will be casily seen, 1 The result tanp tan R corresponds to the fact that the radius of the Nine points circle is half the radius of the circumscribing circle of the triangle. From equation (4) of Art. 144 by supposing the radius of the 6+a sphere to become infinite we obtain i : this corresponds. 2c to the fact that the Nine points circle passes through the feet of the perpendiculars from the angles of a triangle on the opposite sides, From cquation (5)"of Art. 144 by supposing the radius of the sphere to become infinite we obtain u : this corresponds to the 2 fact that the Nine points circle passes through the middle points of the sides of a triangle. From Art. 145 by supposing the radius of the sphere to becoine infinite we obtain %= R cos (A - B): this is a known 2 property of the Nine points circle. In Plane Geometry the points which correspond to the P, G, and N of Art. 146 are on a straight line. 148. The results which have been demonstrated with respect to the circle which touches the inscribed and escribed circles of a spherical triangle are mainly due to Dr Hart and Dr Salmon. See the Quarterly Journal of Mathematics, Vol. vI. page 67. C 1 ·C 112 EXAMPLES. EXAMPLES 2 WE :-) 1, From the formula sin cos S cos (S deduce sin B sin C a sin B sin c the formula for the area of a plane triangle, namely 2 sin A when the tadius of the sphere is indefinitely increased. 2. Two triangles ABC, abc, spherical or plane, equal in all respects, differ slightly in position : shew that cos A Bb cos BCc cos Cua + cos ACccos CBb cos BAa=0. : 3. Deduce formulæ in Plane Trigonometry froun Napier's Analogies. 4. Deduce formulæ in Plane Trigonometry from Delambre's Analogies. C a+b COS sincos A + B 5. From the formula cos deduce 2 2 2 the area of a plane triangle in terms of the sides and one of the angles. 6. What result is obtained from Example 7 to Chapter VI., by supposing, the radius of the sphere infinite ? 7. From the angle of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the sides a and 6: shew that this perpendicular makes an angle S-B with the side ay and an angle S-A with the side b. 8. From each angle of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the adjacent sides. Shew that these perpendiculars meet at a point; and that ·EXAMPLES. 113 if , y, z are the perpendiculars from this point on the sides a, b, c respectively, sin x sin y sin z sin (S - B)sin (3-0) sin (S-C') sin (S - A) sin(S – A) sin(S – B)* 9. Through each angle of a spherical triangle an arc is drawn 80 as to make the same angle with one side which the perpendicular on the base makes with the other side. Shew that these arcs meet at a point; and that if x, y, z are the perpendiculars from this point on the sides a, b, c respectively, sin X sin y sin COS A cos B cos C. 10. Shew that the points determined in Examples 8 and 9, and the point N of Art. 146 are on a great circle. State the corresponding theorem in Plane Geometry. 11. If one angle of a spherical triangle remains constant while the adjacent sides are increased, shew that the area and the sun of the angles are increased. 12. If the arcs bisecting two angles of a spherical triangle and terminated at the opposite sides are equal, the bisected angles will be equal provided their sum be less than 180°. (Let BOD and COE denote these two arcs which are given equal. If the angles B and C are not equal suppose B the greater, Then CD is greater than BE by Art. 58. And as the angle OBC is greater than the angle OCB, therefore OC is greater than OB; therefore 0,D is greater than OE. Hence the angle ODC is greater than the angle OEB, by Example 11. Then construct a spherical triangle BCF on the other side of BC, equal to CBE. Since the angle ODC is greater than the angle OEB, the angle FDC is greater than the angle DFC; therefore CD is less than CF, so that CD is less than BE. See the corresponding problem in Plane Geometry in the Appendix to Euclid, page 317.] T. S. T. I ·( 114 ) XIII. POLYHEDRONS. 149. A polyhedron is a solid bounded by any number of plane rectilineal figures which are called its faces. A polyhedron is said to be regular when its faces are similar and equal regular polygons, and its solid angles equal to one another. 150. If s be the number of solid angles in any polyhedron, F the number of its fuces, E the number of its edges, then S + F= E + 2. Take any point within the polyhedron as centre, and describe a sphere of radius r, and draw straight lines from the centre to each of the angular points of the polyhedron ; 'let the points at which these straight lines meet the surface of the sphere le joined by arcs of great circles, so that the surface of the sphere is divided into as many polygons as the polyhedron has faces. Let 8 denote the sum of the angles of any one of these polygons, m the number of its sides; then the area of the polygon is gol 18 - (m - 2) *} by Art. 99. The sum of the aroas of all the polygons is the surface of the sphere, that is, 47 Hence since the number of the polygons is F, we obtain 4π == Σε – ΤΣm + 2Fπ. Now & denotes the sum of all the angles of the polygons, and is therefore equal to 27 the number of solid angles, that is, to 275'; and Em is equal to the number of all the sides of all the polygons, that is, to 2E, since every edge gives rise to marc which is common to two polygons. Therefore 47 = 273-21E+2F+; therefore S+FE+2.

·POLYHEDRONS. 115 151. There can be only five regular polyhedrons. Let m be the number of sides in each face of a regular polyhedron, n the number of plane angles in each solid angle; then the entire number of plane angles is expressed by mF, or by ns, or by 2E; thus mF=n52E, and S+F= E + 2; from these equations we obtain 4m 2mn 4n SE E 2 (m + n) - mn 2 (m+n) - mn 2 (m + n)- mn These expressions must be positive integers, we must therefore have 2 m + n) greater than mn; therefore 1 1 must be greater than 1 m n 1 n ; ma 1 1 but n capnot be less than 3, so that cannot be greater than 1 1 and therefore must be greater than and as m must be an 6 integer and cannot be less than 3, the only admissible values of mi are 3, 4, 5. It will be found on trial that the only values of m and n which satisfy all the necessary conditions are the following: each regular polyhedron derives its name from the number of its plane faces. nise Name of regular Polyhedron. m 3 3 4 6 3 8 12 3 6 12 4, 3 4 Tetrahedron or regular Pyramid. 6 Hexahedron or Cube. 8 Octahedron. 12 Dodecahedron. 20 Icosahedron. 5 20 30 30 3 5 12 0 It will be seen that the demonstration establishes something more than the enunciation states; for it is not assumed that the faces are equilateral and equiangular and all equal. It is in fact 1 2 ·116 POLYHEDRONS. demonstrated that, there cannot be more than five solids each of which has all its faces with the same number of sides, and all its solid angles formed with the same number of plane angles. 152. The sum of all the plane angles which form the solid angles of any polyhedron is 2 (S-2) 7. For if m denote the number of sides in any face of the polyhedron, the sum of the interior angles of that face is ( m2) * by Euclid 1. 32, Cor. 1. Hence the sum of all the interior angles of all the faces is (m-2)*, that is Em - 2F, that is 2 (E-F), that is 2 (3-2). 153. To find the inclination of two adjacent faces of a regular polyhedron. E Let AB be the edge common to the two adjacent faces, C and D the centres of the faces; bisect AB at E, and join CE and DE; CE and DE will be perpendicular to AB, and the angle CED is the angle of inclination of the two adjacent faces; we shall denote it by I. In the plane containing CE and DE draw CO and DO at right angles to CE and DE respectively, and meeting at 0); about as centre describe a sphere meeting 04, OC, OE at a, c, e respectively, so that cae forms a spherical triangle. Since AB is perpendicular to CE and DE, it is perpendicular to the plane CED, therefore the plane AOB which contains AB is perpendicular to the plane CED; hence the angle cea of the spherical triangle is a right angle. Let m be the number of sides in each face of the "polyhedron, n the number of the plane angles which form each solid ·POLYHEDRONS. 117 27 angle. Then the angle ace = ACE = ;. and the angle 2.m cae is half one of the n equal angles formed on the sphere round az 27 that is, cae = From the right-angled triangle cas cos cae = cos cОe sin ace, TT 2n on 7 T that is COS COS 1) sin n m 7T COS n therefore sin I 2 7T sin m 154. To find the radii of the inscribed and circumscribed spheres of a regular polyhedron. Let the edge AB = a, let OC = q and OA = R, so that r is the radius of the inscribed sphere, and R is the radius of the circumscribel.sphere. Then CE = AE cot ACE = cot m 7T I a g= CE tan CEO=CE tan cot 22 tan I 2 m T also q= R cos aOc= R cot eca cot eac = R cotcot n a therefore R=r tan tan I tan. tan 2 m n n cot m 155. To find the surface and volume of a regulat polyhedron. та? The area of one face of the polyhedron is cot and 4 m ma' therefore the surface of the polyhedron is 4 • Also, the volume of the pyramid which has one face of the ma polyhedron for base and 0 for vertex is cot and 3 4 miral therefore the volume of the polyhedron is eot r . m T 12

·118
POLYHEDRONS.
156. To find the volume of a parallelepiped in terms of its
edges and their inclinations to one another.
B
Let the edges be 0A = a, OB = 6, OC = c; let the inclinations
he BOC=Q, COA=B, AOB=y. Draw CE perpendicular to the
plane AOB meeting it at E. Describe a sphere with O as a
centre, meeting OA, OB, OC, E at a, b, c, e respectively.
The volume of the parallelepiped is equal to the product of its
base and altitude = ab sin y. CE=abc sin y sin cOe. The spherical
triangle cae is right-angled at e; thus
sin coe= sin coa sin cae = sin B sin cab,
and from the spherical triangle cab
sin cab
J(1 - cosa
cos' ß-cosº y + 2 cos a cos ß cos y).
sin ß sin y
therefore the volume of the parallelepiped
=abc (1 – cosa - cos* ß-cosº y + 2 cos a cos ß cos y).
157. To find the diagonal of a parallelepiped in terms of the
three edges which it meets and their inclinations to one another.
Let the edges be 04 = 0, OB=b, OC=c; let the inclinations
be BOC = 4, CO2-B, AOB=y. Let OD be the diagonai iv
quired, and let OE be the diagonal of the face OAB. Then
OD=OES + ED + 20E. ED 008 COE
=a* +6* + 2ab cos y + + 200E cos COE,

·POLYHEDRONS. 119 Describe a sphere with O as centre meeting 04, OB, OC, OE at a, b, c, e respectively; then (see Example 14, Chap. iv.) a B E cos Oe cos cOb sin alle + cos coa sin 60e sin a0b cos a sin a0e + cos B sin b0e sin y therefore 2COE 01) = a +3*ot ce + 2ab cos y + (cos csin a0e + cos ß sin 60e), sin y and O E sin abe=sin y, OE sin 60e = a sin yi therefore ODP= a* +b + O2 + 2ab cos y + 2bc cos a + 2ca cos ß. 158. To find the volume of a tetrahedron. A tetrahedron is one-sixth of a parallelepiped which has the same altitude and its base double that of the tetrahedron; thus if the edges and their inclinations are given we can take one-sixth of the expression for the volume in Art. 156. The volume of a tetrahedron may also be expressed in terms of its six edges; for in the figure of Art. 156 let BC = A', CA-6, AB=d; then 6 + c^_ aº + a'-6" a' +68-62 cos B= 2bc 2ca 2ab and if these values are substituted for cos a, cos B, and cos y in the expression obtained in Art. 156, the volume of the tetrahedron will be expressed in terms of its six edges. The following result will be obtained, in which V denotes the volume of the tetrahedron, COS a = COS Y ·120 POLYHEDRONS. 144 vira" +a'a" (' + * -a) + 6*2" (c* +a" - 6') + cc" (a" + - --a" (a' – 6*) (a' -c) – 6" (6* -c) (6' - a*)-"* (-a") (c" – 6). Thus for a regular tetrahedron we have 144 V = 2a". ‘159. If the vertex of a tetrahedron, be supposed to be situated at any point in the plane of its base, the volume vanishes ; hence if we equate to zero the expression on the right-hand side of the equation just given, we obtain a relation which must hold among the six straight lines which join four points taken arbitrarily in a plane. Or we may adopt Carnot's method, in which this relation is established independently, and the expression for the volume of a tetrahedron is deduced from it; this we shall now shew, and we shall add some other investigations which are also given by Carnot. It will be convenient to alter the notation hitherto used, by interchanging the accented and unaccented letters. . 160. To find the relation holding among the six straight lines which join four points taken arbitrarily in a plano. Let A, B, C, D be the four points. Let AB = C, BC = 0, CA = b; also let DA=a', DB=W, DC = d. If D falls within the triangle ABC, the sum of the angles ADB, BDC, CDA is equal to four right angles; so that cos ADB = cos (BDC + CDA). Hence by ordinary transformations we deduce 1 =cos ADB + cos' BDC + cos' CDA - 2 cos ADB cos BDU COS CDA. If D falls włthout the triangle ABC, one of the three angles at D is equal to the sum of the other two, and the result just given still holds. a +6² - o Now cos ADB and the other cosines may be 2a'b' expressed in a similar manner; substitute those values in the ·POLYHEDRONS. 121 alove result, and we obtain the required relation, which after reduction may be exhibited thus, 0= -a'b + a* a* (6' + ce – ay +6*6* (c+ a' – 69)+c** c' (a’ + b*—c) - a' (a" – b') (a" – c*) – 5 (6" – 6M) (69 - a") -c (cs - a) (c"? b'). 161. To express the volume of a tetrahedron in terms of its six edges. Let a, b, c be the lengths of the sides of a triangle ABC forming one face of the tetrahedron, which we may call its base; let a', 0, d be the lengths of the straight lines which join A, B, C respectively to the vertex of the tetrahedron. Let p be the length of the perpendicular from the vertex on the base; then the lengths of the straight lines drawn from the foot of the perpendicular to A, B, respectively are Jla" - p"), J16"-p), (d-p). Hence, the relation given in Art. 160 will hold if we put /(a" - p)'instead of a', /(b" – p') instead of b', and 'Jlc“ – p') instead of c'. We shall thus obtain p(2a*U* + 20°* + 20'** -as – –c) =- a*b*c* +a"a® (68 + 8 - a*) + b**b* (c* * a' – 6*) + d'%cº (a® + b* -C") -a' (a** – 6") (a" – c) – * (6* – C") (6'* – al) c ('92. a") (e48 – 69). The coefficient of p in this equation is sixteen times the square of the area of the triangle ABC; so that the left-hand member is 144 V, where V denotes the volume of the tetrahedron. Hence the required expression is obtained. 162. To find the relation holding among the six arcs of great circles which join four points taken arbitrarily on the surface of a sphere. Let A, B, C, D be the four points. Let AB = Y, BC = a, CA =ß; let DA=a', DB = B', DC =y'. As in Art. 160 we have 1=cos ADB + cosø BDC + cos' CDA-2 cos ADB cos BDC cos CDA.

·122 POLYHEDRONS. ( cos y - cos a' cos B' Now cos ADB and the other cosines sin a' sin ß' may be expressed in a similar manner; substitute these values in the above result, and we obtain the required relation, which after reduction may be exhibited thus, 1 = cos® a + cos* B + cos” y + cos'' + cos* B' + cosøy - cos a cos a' cos® B cos' B' - cos y cosøy 2Ycos a cos ß cos y + cos a cos ß'cos y + cos B cos a' cos y' + cos y cos a cos B) + 2 (cos a cos ß cos a'cos B' + cos ß cos y cos ß' cos y + cos y cos a cos y cos a'). 163. To find the radius of the sphere circumscribing a tetrahedron. Denote the edges of the tetrahedron as in Art. 1.4k. Let the sphere be supposed to be circumscribed about the tetrahedron, and draw on the sphere the six arcs of greut circles joining the angular points of the tetrahedron. Then the relation given in Art. 162 holds among the cosines of these six arcs. Let , denott the radius of the sphere. Then cox a= 1 – 2 d'9=1--2-1- . 23 and the other cosines may be expressed in a similar manner. Substitute these values in the result of Art. 162, and we obtain, after reduction, with the aid of Art. 161, 4 x 144 V?p= 2a*b*&** + 26*c*%"%d"* + 2c*a'da" - a*a' - 6*6" - c*c". The right-hand member may also be put into factors, as we by recollecting the mode in which the expression for the area of a triangle is put into factors. Let aa'+bb' + cc' = 20; then 36 T?p= 0 (0-aa') (0-66) (o - cc). mee ·EXAMPLES. 123 EXAMPLES. 1. If I denote the inclination of two adjacent faces of a regular polyhedron, shew that cos 1 = $ in the tetrahedron, = 0 in the cube, =- in the octahedron, =- 75 in the dodecahedron, and =- $75 in the icosahedron. 2. With the notation of Art. 153, shew that the radius of the sphere which touches one face of a regular polyhedron and all the adjacent faces produced is fa cot cot I. T m ) 3. A sphere touches one face of a regular tetrahedron and the other three faces produced : find its radius. 4. If a and b are the radii of the spheres inscribed in and described about a regular tetrahedron, shew that b = 3a. 5. Ifach is the radius of a sphere inscribed in a regular tetrahedron, and R the radius of the sphere which touches the edges, shew that R' = 3a. 6. If a is the radius of a sphere inscribed in a regular tetrahedron, and R' the radius of the sphere which touches one face and the others produced, shew that R' = 2a. 7. If a cube and an octahedron be described about a given sphere, the sphere described about these polyhedrons will be the same; and conversely. 8. If a dodecahedron and an icosahedron be described about a given sphere, the sphere described about these polyhedrons will be the same; and conversely. 9. A regular tetrahedron and a regular oetahedron are inscribed in the same sphere : compare the radii of the spheres which can be inscribed in the two solids. 10. The sum of the squares of the four diagonals of a parallelepiped is equal to four times the sum of the squares of the edges.

·124 EXAMPLES 11. If with all the angular points of any parallelepiped as centres equal spheres be described, the sum of the intercepted portions of the parallelepiped will be equal in volume to one of the spheres. 12. A regular octahelron is inscribed in a cube so that the corners of the octahedron are at the cen'cres of the faces of the cube : shew that the volume of the cube is six times that of the octahedron, 13. It is not porsible to fill any given space with a number of regular polyhedrons of the same kind, except cubes; but this may be done by means of tetrahedrons and octahedrons which have equal faces, by using twice as many of the former as of the latter. 14. A spherical triangle is formed on the surface of a sphere 'of radius p; its angular points are joined, forming thus a pyramid with the straight lines joining them with the centre: show that the volume of the pyramid is BpJ(tan r tanr, tans, tan r.), where ?, ?,,?,,r, are the radii of the inscribod and escribed circles of the triangle. 15. The angular points of a regular tetrahedron inscribed in a sphere of radius r being taken as poles, four equal small circles of the sphere are described, so that each circle touches the other threa. Shew that the area of the surface bounded by each circle is my (1-3. 16. If O be any point within a spherical triangle ABC, the product of the sincs of any two sides and the sine of the included angle min 40 sin BO sin Cocot 40 sin BOC - ein 40 sin BO sin co {cot + cot BO sin COA + cot CO sin A AOB ·( 125 ) XIV. ARCS DRAWN TO FIXED POINTS ON TIE „SURFACE OF A SPHERE. 164. In the present Chapter we shall demonstrate various propositions relating to the arcs drawn from any point on the surface of a sphere to certain fixed points on the surface. 165, ABC is a sphoricul triangle having all its sides quadrants, and therefore all its angles right angles; T is any point on the surface of the sphere : to show that cos' TA + cos' TB + cos* TC=1. B T By Art. 37 we have cos 7A = C08 AB cos TB + sin AB sin TB cos TBA sin TB cos TBA. Similarly cos TC=sin TB cos TBC = sin TB sin TBA. Square and add ; thus cos' TA + cos TC=sin' TB=1-008" TB; therefore cos* TA + oos' TB + cos' TC.1.

·126 ARCS DRAWN TO FIXED POINTS 166. ABC is a spherical triangle having all its sides quadrants, and therefore all its angles right angles; T and U are any points on the surface of the sphere : to shew that cos TU = cos TA cos UA+ cos TB cos UB + cos TC cos UC. 7 Com cos TV By Art. 37 we have cos TU - cos 7:1 çox + A +Bin TA sin U A cos TAU, and cos TAU - cos (BAL' - 347) = cos BAU CON BAT + sin BAU sin BAT = cos BA U cos BAT + cos CAUCOS CAT'; therefore = cos TA cos l'A + sin T. sin CA (cos BAU cos BAT + cos CAU cos CAT); and cos T'B == sin TA cos BAT, Cos UB=sin UA COR BAU, cos TC :: stn TA cos CAT, cos UC = sin UA COR CAU; therefore = cos TA cos UA +008 TB COS UB + COB TC COB UC. . cos TU 167. We leave to the student the exercise of showing that the formuke of the two preceding Articles are perfectly general for all ponitions of T. and U, outside or inside the triangle ABC: the ·ON THE SURFACE OF A SPHERE. 127 demonstrations will remain essentially the same for all modifications of the diagrams. The formulæ are of constant application in Analytical Geometry of three dimensions, and are demonstrated in works on that subject; we have given them here as they may be of service in Spherical Trigonometry, and will in fut now be used in obtaining some important results. 168. Let there b? any number of fixed points on the surface of a sphere ; denote them by II, II, I1g.... Let I be any point on the surface of the sphere. We shall now investigate an expression for the sum of the cosines of the arcs which join T with the fixed points. Deuote the sum by S; so that * cos TII, + cos TII, + cos TII, + Tako on the surface of the sphere a fixed spherical triangle ABC, having all its sides quadrants, and therefore all its angles right angles.. Let d, My v be the cosines of the arcs which join T with A, B, C respectively, let l, m, n, be the cosines of the arcs which join II, with A, B, C respectively; and let a similar notation be used with respect to 11, 11,... Then, by Art. 166, S= 1,1 + m, le + nov + led + mgfl + n.3 + ... = PX Qu + Rv; where P stands for 1, +1, +1, + ..., with corresponding meanings for and R. 169. It will be seen that P is the value which takes when T coincides with A, that Q is the value which takes when I' coincides with B, and that R is the value which takes when l'. coincides with C. Hence the result expresses the general value of in terms of the cosines of the arcs which join T to the fixel points A, B, C, and the particular values of which correspond to these three points, ·170. We shall now transform the result of Art. 168. Lat G=1(P+Q+ R"); and let a, B, y le three arcs determined by the equations Р Q R cOS a : COS then = G (1 cos a + he cos B + V cos y). Since cos'a + cos' B + cos'y= 1, it is obvious that there will be some point on the surface of the sphere, such that a, B, y are the arcs which join it to A, B, C respectively; denote this point by U: then, by Art. 166, cos TU= cos a + p cos B + v cos Y ; and finally = G cos TU. Then, whaterer may be the position of T', the sum of the conine's of the arcs which join T to the fixed points varies as the cosine of the single arc which joins T to a certain fixed point We might take G either positive or negative; it will be convenient to suppose it positive. 171. A sphere is described about a regular polyhedron ; from any point on the surface of the sphere ares are drawn to tl. solid angles of the polyhedron: to shew that the sum of the cosines of these arcs is zero. From the preceding Article we see that if G is not zeru there is one position of T which gives to its greatest positive value, namely, when I coincides with l'. But by the symmetry of a regular polyhedron there must always be more than one prosition of T which gives the same value to . For instance, if we take a regular tetrahedron, as there are four faces there will at least be three other positions of T symmetrical with any assigned position, Hence G must be zero; and thus the sum of the cosines of the ATCE schick join T to the solid angles of the reyular polyhedron is zero for all positions of T.

·ON THE SURFACE OF A SPHERE. 129 172. Since G = 0, it follows that P, Q, R must each bo zero; these indeed are peculiar cases of the general result of Art. 171. See Art. 169. 173. The result obtained in Art. 171 may be shewn' to hold also in some other cases. Suppose, for instance, that a rectangulur parallelepiped is inscribed in a sphere, then the sum of the cosines of the arcs druwn from any point on the surfaci of the sphere to the solid angles of the parallelepiped is zero. For here it is obvious that there must always be at least one other position of r syrunetrical with any assigned position. Hence by the urgriment of Art. 171 we must have G=0. 174. Let there be any number of fixed points on the surface of a sphere; denote them by 1, 11, 1,. . Let T be any point on the surface of the sphere. We shall now investigate a remarkable expression for the sum of the squires of the cosines of the arcs which juin 7' with the fixed points. Deuute the sum by 5; so that S = cos TII, + cos® TII, + cos' TH, +.... Take on the surface of the sphere a fixed splerical triangle ABC, having all its sides quadrants, and therefore all its angles right angles. Let ., M, v be the cosines of the arcs which join 7 with A, B, C respectively; let l, m, n, be the cosines of the angles which join II, with A, B, C rospectively; and let a similar notation be used with respect to 11., 1... Then, by Art. 166, =(1,1 + m, pe + nv)' + (1,1 + med + nyv)' + Expand each square, and rearrango the terms; thus S-PX + Qu'+ Rv + 2 puv + 2gva + 2rlyhy where P stands for 1,9 +1,9 + ls' + ..., and p stands for m, n, + m,n+ m, n, + ..., with corresponding meanings for Q and g, and for R and r. T. S. T. + ·130 ARCS DRAWN TO FIXED POINTS We shall now show that there is some position of the triangle ABC for which p, q, and y will vanish; so that we shall then have =PX+ Qu' + Ry'. Since is always a finite positive quantity there must be some position, or some positions, of T for which has the largest valac which it can receive. Suppose that A has this position, or one of these positions if there are more than one. When 7 is at A we have and v each zero, and 2 equal to unity, so that is then equal to P. Hence, whatever be the position of T, P is never less than PX' + Qhea + Ry® + 2 pur + 2qvd + 2rlpe, that is, by Art. 165, P (18+ m + v) is never Jess than PXP + Qu + Rv* + 2 puy + 2q1 +2ru; therefore (P-Q) pe + (P-R) v is never less than puv + 2qvd + 2rdpl. Now suppose v= = 0; then T is situated on the great circle of which AB is a quadrant, and whatever be the position of T we have (P-Q) not less than 2rly, 2rà and therefore P-Q not less than 1 cos TA 1 λ But now is equal to cos T'B; this is numerically equal to р tan TB, and so may be made numerically as great as we please, positive or negative, by giving a suitable position to T. Thus 2ra P-must in some cases be less than if y have any value dif- M ferent from zero. Therefore r must 0. In like manner we can show that a must = 0.

·ON THE SURFACE OF A SPHERE. 131 Hence with the specified position for A we arrive at the result that whatever may be the position of T S=PX' + Qu+ Rv® + 2puv. Let us now suppose that the position of B is so taken that when T coincides with B the value of is as large as it can be for any point in the great circle of which A is the pole. When | is at B we have 1 and v each zero, and y equal to unity, so that is then equal to Q. For any point in the great circle of which A is the pole 1 is zero; and therefore for any such point Q is not less than Qu' + Rv + puv, that is, by Art. 105, Q (+ v) is not less than Qja + Rve + ?puv; 2pu therefore Q-R is not less than Hence by the same reasoning as before we must have p=0. Therefore we see that there must be some position of the triangle ABC, such that for every position of T' •=PX + Qu + Rvº. 175. The remarks of Art. 169 are applicable to the result just obtained. 176. In the final result of Art. 174 we may shew that R is the least value which can receive. For, by Art. 165, SPX' + Qu'+R(1-1-M") = R +(P-R) X+(Q-R)p'; and by supposition neither P-R nor Q-R is negative, so that cannot be less than R. • 77.. A sphere is described about a regular polyhedron ; from any point on the surface of the sphere arcs are drawn to the solid angles of the polyhedron : it is required to find the sum of the squares of the cosines of these arcs. K2 ·132 ARCS DRAWN TO FIXED POINTS With the notation of Art. 174 we have = P + Qu + Rv. We shall shew that in the present case P, Q, and R must all be equal. For if they are not, one of them must be greater than each of the others, or one of them must be less than cach of the others. If possible let the former be the case; suppose that P is greater than 6, and greater than R. Now =P(1 -'-1) + Qu' + Rv* =P-(P-u - (P-R) v* ; this shews that is always less than P except when u = 0 and v=0 : that is is always less than P except when I' is at A, or at the point of the surface wbich is diametrically opposite to A. But by the symmetry of a regular polyhedron there must always be more than two positions of T which give the same value to .. For instance if we take a regular tetrahedron, as there are four faces there will be at least three other positions of T'nsymmetrical with any assigned position. Hence P call not be greater than Q and greater than R. In the san'e way we can shew that one of the three P, Q, and R, cannot be less than each of the others. Therefore P=Q=R; and therefore by Art. 165 for every position of T we have = P. Since P=Q= R each of them = }(P+Q+R) {* + m^+ % * +++ m + + ...} -, by Art. 165, where & is the number of the solid angles of the regular polyhedron.

·ON THE SURFACE OF A SPHERE. 133 Thus the sum of the squares of the cosines of the arcs which join any point on the surface of the sphere to the solid angles of the regular polyhedron is one third of the number of the solid angles. 178. Since P=Q=R in the preceding Article, it will follow that when the fixed points of Art. 174 are the solid angles of a regular polyhedron, then for any position of the spherical triungle ABC we shall have p=0, q = 0, and r=0. For taking any position for the spherical triangle ABC we have = PA + Qu* + Rv + 2 puv + 2qvd + 2rdu; then at A wo have j = 0 and v=0, so that P is then the value of ; siruilarly Q and R are the values of at B and C respectively. But by Art. 177 we have the same value for whatever be the position of T;. thus Po=P(1* + * + v*) + 2 puv + 2q101 + 2rlu; therefore 0 = 2ppv + 2qva + 2 rdquo This holds then for every position of T. Suppose T is at any point of the great circle of which A is the pole; then .= 0 : thus was get puv = 0, and therefore p=0. Similarly q = 0, and r = 0. 179. Let there be any number of fixed points on the surface of a sphere ; dcnote them by II, II,, II, ...; from any two points T and U on the surface of the sphere arcs are drawn to the fixed points : it is required to find the sum of the products of the corresponding cosines, that is cos TII,'cos UH, + cos TH cos UH, + cos TA, cos UH. + ... Let the notation be the same as in Art. 174 ; and let X', u', v be the cosines of the arcs which join U with A, B, C respectively. Then by Art. 166, cos TH, cos UH, = (4x + um, +mn) (X7, + j'm, + v'n) 117,' + pepe'm, ' +w'n, * +(aje'+rell)2,00, +(rev' +vu') mini+(vx'+do) nelze 1 ·134 ARCS DRAWN TO FIXED POINTS. Similar results hold for cos TH, cos UH,, cos TH, cos VII,, ... Hence, with the notation of Art. 174, the required sum is WP+Mpe'Q + w'R+(ur' + vp)p+ (vx' + dv) 9+ (Ape' +pid 9 r. Now by properly choosing the position of the trianglo ABC we have p, q, and each zero as in Art. 174; and thus the required sum becomes MP + up'Q+w'R. 180. The result obtained in Art. 174 may be considered as a particular case of that just given; namely the case in which the points T and U coincide. 181. A sphere is described about a regular polyhedron; from any two points on the surface of the splicre arcs are drawn to the solid angles of the polyhedren: it is required to find the sum of the products of the corresponding cosines. With the notation of Art. 179 we see that the sun 18 λXP + Men'Q + w'R. S And here P=Q=R by Art. 177. 3' S Thus the sun = ('+Me' + rv) =- S 3 cos TU. Thus the sum of the products of the cosines is equal to the product of the cosine of TU into a third of the number. of the solid angles of the regular polyhedron. 182. The result obtained in Art. 177 may be considered as a particular case of that just given; namely, the case in which the points T and U coincide. 183. If TU is a quadrant then cos TU is zero, and the sum of the products of the cosines in Art. 181 is zero. The results p= 0, q=0, r=0, are easily seen to be all special examples of this particular case.

·'( 135 ) XV. MISCELLANEOUS PROPOSITIONS. 184. To find the locus of the vertex of a spherical triangle of given base and areas Let AB be the given base, = c suppose, AC = 0; BAC = $. Since the area is given the spherical excess is known; denote it by E; then by Art. 103, cot : E = cot 10 cot } c cosec $ + cotº; therefore sin (-E)=cot 10 cot c sin } E; A therefore 2 cotc sin } E cos = sin sin ($- $ E); 2 therefore cos 6 cot | c sin. E + sin 8 cos($- E+ 122) -cot c sin & E. Comparing this with equation (1) of Art. 133, we see that the required locus is a circle. If we call a, ß the angular co-ordinates of its pole, we have tan c. tan a= 1 cot $c sin E sin' 7T B = 1 E It may be presumed from symmetry that the pole of this circle is in the great circle which bisects AB at right angles ; and this presumption is easily verified. For the equation to that great circle is 0 = cos A cos + sin sin ) 2 cos (6-6) and the values 0 = 0, $=B satisfy this equation, -j) N ·136 MISCELLANEOUS PROPOSITIONS. 185. To find the angular distance between the poles of the inscribed and circumscribed circles of a triangle, Let P denote the pole of the inscribed circle, and the pole of the circumscribed circle of a triangle ABC; then PAB= $ A, by Art. 89, and QABS-C, by Art. 92, hende cos PAQ = cos (B-C); and cos PQ=cos PA cos Q4 + sin PA sin QA cos : (B-C). Now, Wy Art. 62 (see the figure of Art. 89), = cos PE cos A E = cos r cos (8 - a), sin PE sin PA = sin PAE sinA thus cos PA sin r G cos PQ=cos R cos a cos (8 - a) +sin R sin r cos } (B-C) cosec ! A. Therefore, by Art. 54 cos PQ = cos R cos r cos (8 - a) + sin R sin r sin } (b + c) cosec 1 n cos PQ therefore cotr cos (8-- a) + tan R'sin } (b+c) cosec f a. cos R sinr ** sin 8 Now 2 sin f a sin f b sin fc, cotrs tan R n cos PQ therefore cos R sinr 11 sin 8 cos (8-a) + 2 sin } (b+c) sin f 6 sin in to n 1 (sin a + sin b + sin c). 2n Hence 1 (sin a + sin b+ sin c)-1 4n' therefore (oot r+tan R)* (by Art. 94); cos' PQ = cos' R sinr + cos* (R-r), sin' PQ =sin' (R-1) - cos' R sinór. and ·MISCELLANEOUS PROPOSITIONS. 137 186. To find the angular distance between the pole of the circumscribed circle and the pole of one of the escribed circles of a triangle. Let Q denote the pole of the circumscribed circle, and Q, the pole of the escribed circle opposite to the angle A. Then it may be shewn that QBQ,= 17+ 4 (C-4), and cos QQ, =cos R cos r, cos (8 –c) - sin R sin r, sin } (C - A) sec B = cos R cos r, cos (s—c) -- sin R sin r, sin 1 (e-a) cosec lb. Therefore cos QQ, =cot r, cos (8-c) – tan R sin } (c-a) cosec 1 b; sin r, cos R by reducing as in the preceding Article, the right-hand member of the last equation becomes 1 hence coses hora) - 1 = (tan R-cot r.)(Art. 94); (sin b + sin c- sin a); 272 cos QQ cos R sinar COS' QQ,=cos' R sinºr, + cos' (R+r), sin' QQ2 = sin(R+ r) - cos' R sin r.. therefore and . 187. The arc which passes through the middle points of the sides of any triangle upon a given base will meet the base produced at a fixed point, the distance of which from the middle point of the base is a quadrant. Let ABC be any triangle, E the middle point of AC, and F the middld point of AB; let the arc which joins E and F when produced meet BC produced at Q. Then sin BQ sin BFQ sin AQ sin AFQ sin BQF' sin AQF sin BQ sin AQF therefore sin AQ sin BQF' sin BF sin AF ·138 MISCELLANEOUS PROPOSITIONS. sin CQ sin AQF similarly sin 4Q sin CFI therefore sin BQ=sin CQ; therefore BQ + CQ = 7. Hence if D be the middle point of BC DQ= $ (BQ + C) = } . 188. If three arcs be drawn from the angles of a spherical triangle through any point to meet the opposite sides, the products of the sines of the alternate segments of the sides are equal. F P Let P be any point, and let arcs be drawn from the angles A, B, C passing through P and meeting the opposite sides at D, E, F. Then sin BD sin BPD sin CD sin CPD, sin BP sin BDP sin CP sin CDP sin BD sin BPD sin BP therefore sin CD sin CPD sin CP. sin CE sin AF Similar expressions may be found for and sin AE and hence it follows obviously that sin BF 1; sin BD sin CE sin AF sin CD sin Ať sin BF therefore sin BD sin CE sin AP= sin CD sin AE sin BF. f.

·MISCELLANEOUS PROPOSITIONS. 139 189. Conversely, when the points D, E, F in the sides of a spherical triangle are such that the relation given in the preceding Article holds, the arcs which join these points with the opposite angles respectively pass through a common point. Hence the following propositions may be established: the perpendiculars from the angles of a spherical triangle on the opposite sides meet at a point; the arcs which bisect the angles of a spherical triangle meet at a point; the arcs which join the angles of a spherical triangle with the middle points of the opposite sides meet at a point; the arcs which join the angles of a spherical triangle with the points where the inscribed circle touches the opposite sides respectively meet at a point. Another mode of establishing such propositions has been exemplified in Arts. 139 and 140. 190. If AB and A'B' be any two equal arcs, and the arcs, AA' and BB' be bisected at right angles by arcs meeting at P, then AB and A'B' subtend equal angles at P. P .............1 B For PA = PA' and PB=PB'; hence the sides of the triangle PAB are respectively equal to those of PA'B'; therefore the angle APB = the angle A'PB. This simple proposition has an important application to the motion of a rigid body of which one point is fixed. For conceive a sphere capable of motion round its centre which is fixed; then it appears from this proposition that any two fixed points on the ·140 MISCELLANEOUS PROPOSITIONS. sphere, as A and B, can be brought into any other positions, as A' and B, by rotation round an axis passing through the centre of the sphere and a certain point P. Hence it may be inferred that any change of position in a rigid body, of which one point is fixel, may be effected by rotation round some axis through the fixed point. (De Morgan's Differential and Integral Calculus, page 489.) 191. Let P, denote any point within any plane angle AOB, and from P draw perpendiculars on the straight lines Od and OB; then it is evident that these perpendiculars include an angle which is the supplement of the angle 40B. The corresponding fact with respect to a solid angle is worthy of notice. Let there be a solid angle formed by three plane angles, meeting at a point 0. From any point P within the solid angle, draw perpendiculars PL, PM, PV on the three planes which form the solid angle; then the spherical triangle which corresponds to the three planes LPM, MPY, NPL is the polar triangle of the spherical triangle which corresponıls to the solid angle at 0. This reinark. is due to Professor De Morgan. 192. Suppose three straight lines to meet at a point and form a solid angle ; let a, b, and y denote the angles contained by these three straight lines taken in pairs: then it has been proposed to call the expression (l-cos'a - cos'ß- cos'y + cos a cos ß cos y), the sine of the solid angle. See Baltzer's Theorie...der Determinanten, 2nd edition, page 177. Adopting this definition it is easy to shew that the sine of a solid angle lies between zero and unity. We know that the area of a plane triangle is half the product of two sides into the sine of the included angle : by Art. 156 we have the following analogous proposition; the volume of a tetrahedron is one sixth of the product of three edges into the sine of the solid angle which they form. Again, we know in mechanics that if three forces acting at a point are in equilibrium, each force is as the sine of the angle between the directions of the other two: the following proposition is analogous; if four forces acting at a point are in equilibrium ·MISCELLANEOUS PROPOSITIONS. 141 each force is as the sine of the solid angle formed by the directions of the other three. See Statics, Chapter II. 193. Let a sphere be described about a regular polyhedron; let perpendiculars be drawn from the centre of the sphere on the faces of the polyhedron, and produced to meet the surface of the sphere : then it is obvious from symmetry that the points of intersection must be the angular points of another regular polyhedron. This may be verified. It will be found on examination that if S be the number of solid angles, and F the number of faces of one regular polyhedron, then another regular polyhedron exists which has S faces and F solid angles. See Art. 151. 194. Polyhedrons. The result in Art. 150 was first obtained by Euler; the demonstration which is there given is due to Legendre. The demonstration shews that the result is true in many cioses in which the polyhedron las re-entrant solid angles," for all "hat is necessary for the demonstration is, that it shall be possible to take a point within the polyhedron as the centre of a sphere, so that the polygons, formed as in Art. 150, shall not have any coincident portions. The result, however, is generally true, cases in which the condition required by the demonstration of Art. 150 is not satisfied. We shall accordingly give another demonstration, and shall then deduce some important consequences from the result. We begin with a theorem which is due to Cauchy. 195. Let there be any network of rectilineal figures, not necessarily in one plane, but not forming a closed surface; let E be the number of edges, F the number of figures, and S the number of corner points: then F+S=E+1. This theorem is obviously true in the case of a single plane figure; for then F=1, and S= E. It can be shown to be generally true by induction. For assume the theorem to be true for a network of F figures; and suppose that a rectilineal figure of n sides is added to this network, so that the network and the additional figure have m sides coincidont, and therefore m +1 even ·142 MISCELLANEOUS PROPOSITIONS. corner points coincident. And with respect to the new network which is thus formed, let E, F, S denote the same things as E, F, S with respect to the old network. Then E' = E + n - m, F=F+1, S=S+n-(m + 1); therefore F+S-E=F+S-E. But F+SE+1, by hypothesis; therefore 1" + S'= E +1, 196. To demonstrate Enler's theorem we suppose one face of a polyhedron removed, and we thus obtain a network of rectilineal figures to which (auchy's theorem is applicable. Thus F-1+S=E+1; therefore F+S=B+2. 197. In any polyhedron the number of faces with an odd number of sides is even, and the number of solid angles formed with an odd number of plane angles is even. Let a, b, c, d,...... denote respectively the numbers of faces which are triangles, quadrilaterals, pentagons, lexigons,...... Let a, B, 7, 8, ...... denote respectively: the numbers of thie solid angles which are formed with three, four, five, six,...... plane angles. Then, each edge belongs to two fuces, and terminates at two solid angles; therefore 27 = 3a + 4b + 5c + Od +.. 2E = 3a +43 + 5y + 68 +....... From these relations it follows that a+c+e+ and a+y+ e +...... are even numbers. 198. With the notation of the preceding Article we have F=a + b + c + d +...... S=Q+B+y+ 8 +...... From these combined with the former relations we obtain 2E - 37=6 + 2c + 3d + 2E - 3S=B+ 2y + 38+ ...... Thus 2E cannot be less than 37, or less than 3S. .....

·MISCELLANEOUS PROPOSITIONS. 140 199. From the expressions for E, F, and S, given in the two preceding Articles, combined with the result 2F+25=4+2E, we obtain 2(a+b+c+d+ ...)+2(a+B+y++...)= 4 + 3a + 4b + 50+6d+..., 2(a+b+c+d+..:)+2(a +B+y++...)=4 + 3a +4% + 5y + 68 +..., thereforo 2 (a + ß+y+8+ ...) - (a + 25 + 3c + 4d + ...)=4...(1), 2 (a+b+c+d+ ...) - (a + 2B + 3y +48 + ...)= 4...(2). Therefore, by addition a +a -(c+ y) - 2 (d+ 8) - 3 (e + e) –...... = 8. Thus the number of triangular faces together with the number of solid angles formeil with three plane angles cannot be less than eight. Again, froin (1) and (2), by climinating a, we obtain 3a + 2b +c-e-?f - ... - 2B – 4y - 12, so that ia +26+c cannot be less than 12. From this result various inferences can be drawn; thus for example, a solid cannot be formed which shall lave no triangular, quadrilateral, or pentagonul faces. In like manner, we can show that 3a + 2B + y cannot be less than 12. ...... 200. Poinsot has shewn that in addition to the five wellknown regular polyhedrons, four other solids exist which are perfectly symnietrical in shape, and which might therefore also be called regular. We may give an idea of the nature of Poinsot's results by referring to the case of a polygon. Suppose five points A, B, C, D, E, placed in succession at equal distances round the circumference of a circle. If we draw a straight line from each point to the next point, we form an ordinary regular pentagon, Suppose however we join the points by straight lines in the following order, A to C, C to E, E to B, B to D, D to A ; we thus form a star-shaped symmetrical figure, which might be considered a regular pentagon.

·MISCELLANEOUS PROPOSITIONS. It appears that, in a like manner, four, and only four, new regular solids can be formed. To such solids, the faces of which intersect and cross, Euler's theorem does not apply. 201. Let us return to Art. 195, and supposo e the number of edges in the bounding contour, and é the number of edges within it; also suppose : the number of corners in the bounding contour, and s' the number within it. Then E=e+e'; $= 8 +8'; therefore 1 + e +é = 8 + 8 + F. But therefore 1+ d = 8' + F. We can now demonstrate an extension of Euler's theorem, which has been given by Cauchy. 202. Let a polyhedron be decomposcul into any number of polyhedrons at pleasure; let P be the number thus formed, s the number of solid angles, F the number of faccs, E the number of edges: then S + F = E + P + 1. For suppose all the polyhedrons uniterl, by starting with one and adding one at a time. Let e, f, 8 be respectively the numbers of edges, faces, and solid angles in the first; let é, J", ' be respectively the numbers of edges, faces, and solid angles in the second which are not common to it and the first; let e", S", 8" be respectively the numbers of edges, faces, and solid angles in the third which are not common to it and the first or second; and so on. Then we have the following results, namely, the first by Art. 196, and the others by Art, 201; 8 + f = e + 2, $+" ad + 1, 8"+"="+1, o ...... By addition, since 8 + 8' + 8" + = S, f + g + y + ... = F, and e +ég+t" + = E, we obtain 8+. E + P + 1.

·MISCELLANEOUS EXAMPLES. 145 203. The following references will be useful to those who study the theory of polyhedrons. Euler, Novi Commentarii Academie.... Petropolitana, Vol. iv. 1758; Legendre, Géométrie; Poinsot, Journal de l'Ecole Polytechnique, Cahier x; Cauchy, Journal de l'Ecole Polytechnique, Cahier XVI; Poinsot ad Bertrand, Comptes Rendus ... de l'Académie des Sciences, Vol. XLVI; Catalan, Théorèmes et Problèmes de Géométrie Elémentaire ; Kirkman, Philosophical Transactions for 1856 and subsequent years ; Listing, Abhandlungen der Königlichen Gesellschaft... Gortingen, Vol. x. MISCELLANEOUS EXAMPLES. 1. Find the locus of the vertices of all right-angled spherical triangles having the same hypotenuse; and from the equation obtained, prove that the locus is a circle when the radius of the sphere is infmite. 2. AB is an arc of a great circle on the surface of a sphere, C its middle point : show that the locus of the point P, such that the angle A PC = the angle BPC, consists of two great circles at fight angles to one another. Explain this when the triangle becomes plunc. 3. On a given arc of a sphere, spherical triangles of equal area are described : shew that the locus of the angular point opposite to the given arc is defined by the equation tan (a + 0) tan (a -6) sin tan *) } + tan sin + tan tan o sin (a + o)) 5} + tantan a sin (a-0) =B, where 2a is the length of the given arc, 0 the arc of the great circle drawn from any point P in the locus perpendicular to the given arc, $ the inclination of the great circle on which is T. S. T. L ·146 MISCELLANEOUS EXAMPLES meusured to the great circle bisecting the given arc at right angles, and ß & constant. 4. In any spherical triangle cot A cot a tan c=-- cot a cot 6 cot B coto cos i cos B 5. If 0, 0, y denote the distances from the angles A, B, C respectively of the point of intersection of arcs bisoeting the angles of the spherical triangle ABC, shew that cos O sin (6 - c) + cos o sin (c -- a) + cos sin (a - b) = 0. 6. If 4', B', C'he the poles of the sides Br, CA, 4 B of a spherical triangle ABC, shew that the great circles 11', BB, CC meet at a pint P, such that cos P. cos BC = cos P'B COXCA.cos PC cos AB. 7. If o be the point of intersection of arcs AD, BE, CH' drawn from the angles of a triangle purpendicular to the opposite sides and meeting them at D, E, F' respectively, show that tan AD tan BE tan CF till OD' tan OE' tan 01 are respectively equal to cos B cos C 1+ cos A cos B cos C 1 + 1 + cos A cos C cos d cos B' 8. If p, q, qo be the arcs of great circles drawn from the angles of a triangle perpendicular to the opposite sides, (a, a), (B, B), (y, the segments into which these arcs are divided, show that tan a tan a' = tan ſ tan B' = tan y tan y'; COSP co89 COST and cos a cos a cos ß cos ßcos y cos y ·MISCELLANEOUS EXAMPLES. 147 9. In a spherical triangle if arcs be drawn from the angles to the middle points of the opposite sides, and if an a' be the two parts of tho one which bisects the side a, shew that sin a a 2 cos sin a 2 10. The arc of a great circle bisecting the sides AB, AC of a spherical triangle cuts BC produced to: shew that a C 6 C + 6 cos AQ sin sin sin 2 2 11. If ADCD be a spherical quadrilateral, and the opposite sides A1, CD when procluced meet at E, and AD, BC meet at F, the ratio of the sincs of the arcs drawn from E at right angles to the diag muls of the juadrilateral.is the same as the ratio of those from 7 12.. I ABCD) be a spherical quadrilateral whose sides AB, DC are produced to meet at P, and AD, BC at Q, and whose diagonals AC, BD intersect at R, then sin AB sin CD cos P - sin AD sin BC cos Q sin AC sin BD cos R. 13. If A' be the angle of the chordal triangle which corresponds to the angle A of a spherical twiangle, shew that a cos A' =sin (S - A) cos 2 14. If the tangent of the radius of the circle described about a spherical trianglo is equal to twice the tangent of the radius of the circle inscribed in the triangle, the triangle is equilateral. 15. The arc AP of a circle of the same radius as the sphere is equal to the greater of two sides of a spherical triangle, and the arc AQ taken in the same direction is equal to the less; the EM sine PM of AP is divided at E, so that the natural cosine PM L2 ·148 MISCELLANEOUS EXAMPLES. of the angle included by the two sides, and EZ is drawn parallel to the tangent to the circle at Q. Shew that the remaining side of the spherical triangle is equal to the arc QPZ. 16. If through any point P within a splerical triangle ABC great circles be drawn from the angular points A, B, C to meet the opposite sides at a, b, c respectively, prove that sin Pa cos PA sin Pb cos PB sin Pc cos PC -1. sin B6 sin Cc + sin da 17. A and B are two places on the Earth's surface on the same side of the equator, A being further from the equator than B. If the bearing of A from B be more nearly due East than it is from any other place in the same latitude as B, what is the bearing of B from A? 18. From the result given in example 18 of Chapter v. infor the possibility of a regular dodecahedron. 19. A and B are fixed points on the sur ce of a sphere, and P is any point on the surface. If a and b are given constants, shew that a fixed point S can always be found, in AB or AB produced, such that a cos AP + b cos BP = 8 cos SP, where : is a constant. 20. A, E, C,... are fixed points on the surface of a sphere; a, b, c,... are given constants. If P be a point on the surface of the sphere, such that a cos AP + 6 cos BP + c cos CP + constant, shew that the locus of P is a circle. ...

·( 149 ) XVI. NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 204. We shall give in this Chapter examples of the numerical solution of Spherical Triangles. We shall first tako right-angled triangles, and then obliqueangled triangles. Right-Angled Triangles. 205. Given a=37° 48' 12", b = 59° 44' 16", C= 90°. To find c we have cos c= cos a cos 6, L cos 37° 48' 12" 9.8976927 I cos 59° 44' 16" - = 9.7023945 I cos c+ 10 = 19.6000872 66° 32' 6", To find A we have cot A =cot a sin b, I cot 37° 48' 12' = 10.1102655 L sin 59° 44' 16" = 9.9363770 L cot A + 10 = 20.0466425 A = 41° 55' 45".

·150 NUMERACAL SOLUTION OF SPHERICAL TRIANGLES.
To find B we have
pot B=cot b sin
a,
L cot 59° 44' 16"
L sin 37° 48' 12"
9-7660175
9.7874272
L cot B +10= 19.5534447
B-70° 19' 15".
206. Given A= 55° 32 45", C = 90°, c-: 98° 14' 24".
To find a we have
sin a= sin c sin A,
L sin 98° 14' 24" = 9.9954932
L sin 55° 32' 45" = 9.9162323
;
L sin a + 10 = 19.9117255
a = 54° 41' 35".
To find B we have
cot B - = cos c tan 4.
Here cosc is negative; and therefore cot B will be negative,
and B greater than a right angle. The numerical value of cos c
is the same as that of eos 81° 45' 36".
I cos 81° 45' 36" : 9.1563065
L tan 55° 32' 45' = 10.1636102
L cot (180º – B) + 10 = 19.3199167
180° - B= 78° 12' 4"
B = 101° 47' 56".

·NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 151
To find b we have
tan b = tan c cos A.
Here tan c is negative ; and therefore tan b will he negative
and b greater than a quadrant.
L tan 81° 45' 36' = 10.8391867
L cos 55° 32' 45' == 9.7526221
L tan (180° – 6) + 10 = 20-5918088
180° - 6 = 75° 38' 32"
b = 104° 21' 28".
207. Given A = 46° 15' 25", C = 90°, a = 42° 18' 45".
To find c we have
sin a
sin c=
sin A'
L Sin c= 10 + L sin a-L sin A,
10 + L sin 42° 18' 45" = 19.8281272
L sin 46° 15' 25" = 9.8588065
L sin c = 9.9693207
68° 42' 59" or 111' 17' 1".
To find 6 we have
sin b = tan a cot A,
9.9591983
L tan 42° 18' 45".
L cot 46° 15' 25":
9.9809389
L sin b + 10 = 19.9401372
b = 60° 36' 10" or 119° 23' 50%.

·152 NUMERICAL SOLUTION OF SPHERICAL TRIANGLES., To find B we have CO8 A sin B COS a L sin B = 10 + L cos A-L cos a, 10 + L cos 46° 15' 25" = 19.8397454 L cos 42° 18' 45" 9.8689289 L sin B 9.9708165 *B=69° 13' 47" or 110° 46' 13". Oblique-Angled Triangles. 38° 46' 10". 208. Given a = 70° 14' 20", b= 49° 21' 10", c We shall use the formula given in Art. 45, tan ļ4 = {{sin 18 – 6) sin (8-01 - a a) sin 8 sin ( 8 79° 12' 20", Here 8° 58', . 3-6 = 29° 48' 10", 8-C= 40° 26' 10". L sin 29° 48' 10" = 9.6963704 L sin 40° 26' 10" 9.8119768 19.5083472 L sin 79° 12' 20%= 9.9922465 L sin 8° 58' = 9.1927342 19.1849807 19.5083472 19.1849807 2) 3233665 L tan 4 - 10 = •1616832 *4 = 55° 25' 38" A = 110° 51' 16".

·NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 153 Similarly to find B, L sin 80 58 9.1927342 L sin 40° 26' 10" = 9.8119768 19.0047110 i sin 79° 12' 20" L sin 29° 48' 10": 9.9922465 9.6963704 19-6886169 19.0047110 19.6886169 2) 1.3160941 L tan \'B-10 = 1.6580470 L tan B: 9.6580170 *B= 24° 28' 2" B=48° 56' 4". Similarly to find C, I sin 8° 58' - 9.1927342 L sin 29° 48' 10" = 9:6963704 18.8891046 L sin 790 12' 20" L sin 40° 26' 10". 9.9922465 9.8119768 19.8042233 18.8891046 19.8042233 2) 1.0848813 I tan 10-10= 1.5424406 L tan C = 9.5424406 BC = 19° 13' 24" C = 38° 26' 48".

·15+ NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 209. Given a = 68° 20' 25", b=52° 18' 15", C=117° 12' 20". By Art. 82, tan (A + B) = cos } (a - b) cot! C, cos } (a+b) sin (a-6) tan (A - B) = cot : C. sin } (a + b) 1 (a - b) = 8° 7'5", }(a+b) = 60° 19' 20", C = 58° 36' 10". L cos 8" 1' 5" 9.9957331 L cot 58° 36' 10" = 9.7855690 197813025 I cos 60° 19' 20". 9.6947120 L tan ( 1 + B) = 10:03G5905 (A + B) = 50" 40' 28" L sin 8° 1' 5" 9.1445280 L cot 58° 36' 10" = 9.7855690 18.9300970 9.9389316 L sin 60° 19' 20". L tan } (A - B)= 8.9911654 (A - B) = 5° 35' 47". Therefore A = 56° 16' 15", B = 45° 4'41". If we proceed to find c from the formula sin a sin C ginc sin A since sin C is greater than sin A we shall obtain two values for c both greater than a, and we shall not know which is the value to ·. NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 155 We shall therefore determine c from formula (1) of Art. 54, which is free from ambiguity, > cos cos } (a + b) sin 1C cos ) (A + B) I cos 60° 19' 20"= 9.6947120 I sin 58° 36' 10" 9.9312422 19.6259542 L cos 50° 40' 28" = 9.8019015 L cosc 9.8240527 1 c = 48° 10' 22" c= 96° 20' 44". Or we may adopt the second method of Art. 82. First, we determino o tron the formula tan 0= tan b cos C. Here cos C is negalive, and therefore tan 8 will be negative, and a greater than a right angle. The numerical value of cos is the sanie as that of cos 620047' 40". L tin 52° 18 15” = 10:1119488 L cos 62° 4" 40": 9.6600912 L tan (180°-6) + 10 = 19.7730400. 180° - 0= 30' 36' 33", therefore A = 149° 23' 37". Next, we determine c from the formula COS C cos 6 cos ( (a-6) cos Here cos 0 is negative, and therefore cosc will be negative, and 'c will be greater than a right angle. The numerical value of cos 0 is the same as that of cos (180° – 0), that is, of cos 30° 36' 33"; and the value of cos (a - b) is the same as that of cos (0 - a), that is, of cos 81° 3' 2".

·156 NUBIERICAL SOLUTION OF SPHERICAL TRIANGLES. I cos 52° 18' 15" L cos 81° 3' 2" 978637 48 9.1919060 18.9782808 L cos 30° 36' 33": 9.9348319 I cos (180°-C) - 9.0434489 180° - C = 83° 39' 17" c= 96° 20' 43". Thus by taking only the nearest number of seconds in the tables the two methods give values of c which differ by l"; if, however, we estimate fractions of a second both methods will agree in giving about 434 as the number of seconds. 210. Given a = 50° 45' 20", b = 69° 12' 40", 4 = 44° 22' 10". By Art. 84, sin B sin 6 sin a sin A, L sin 69° 12' 40" 9.9707626 L sin 44° 22' 10"= 9.8446.525 19.8154151 L sin 50° 45' 20" :- 9.8889956 L sin B= 9.9264195 B = 57° 34' 51":4, or 122° 25'8".6. In this case there will be two solutions; see Art. 86. Wo will calculate C and c by Napier's analogies, cos (b-a) tan cot 1 (B+ A), cos * (6 + a) cos 1 (B + A) tan 1 (b+a). cos 1 (B-A) First take the smaller value of B ; thus 1 (B + A)= 50° 58' 30":7, (B- A) = 6° 36' 20":7, tan ·NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 157 L cos 9° 13' 40" 9.9943430 L cot 50° 58' 30":7 = 9.9087536 = 19.9030966 L cos 59° 59' = 9.6991887 L tan! C - 10.2039079 ! C= 57° 58' 55".3 C-115° 57' 50".6. L cos 50° 58' 30":7 = 9.7991039 L tan 59° 59' = 10.2382689 20.0373728 L cos 6° 36' 20".7- 9.9971072 I tan : . c=10.0102656 C == 47° 39' 8":2 c=95° 18' 16"4. Next Take the larger value of B ; thus (B + A) -- 83° 23'039":3, (B- A) -- 39° 1' 29"-3. L cos 9° 13' 40" 9.3943430 L cot 83° 23' 39".3 : 9:0637297 19.0580727 9.6991887 L cos 59° 59' = L tan} C= 9:3588840 C= 12° 52' 15":8 C= 25° 44' 31".6. L cos 83° 23' 39":3= 9.0608369 L tan 59° 59' - 10.2382689 19.2991058 L cos 39° 1' 29":3= 9.8903494 L tan c= 9:4087564 3c=14° 22' 32":6 28° 45' 5".2 ·158 EXAMPLES The student can obtain more examples, which can be easily verified, from those here worked out, by interchanging the given and required quantitics, or by muking use of the polar triangle. EXAMPLES. 1. Given b = 137° 3' 48", A= 1.170 % 54", C. 90". Results. 47° 57' 15", a = 156° 10' 34", B = 113° 28'. 2. Given 61" 4'56”, a -- 40° 31' 20', C = 30'. Results. b= 50° 30' 29", B = 61' 50' 28", A - 47° 54' 21", CE- 3. Given A = 36', B = 60', C-: 90'. Results. = 20°54' 18":7, O 31" 43' 3", c.37" 21' 38".). с 138' 4", 4. Given a = 59° 28' 27", 1:'66" 7' 20", C=!0. Results. 70° 23' 42", 6 - 48° 39' 16", B = 52" 50' 20", or, 109° 30' 18", b=131° 20' 41", B = 127.9'40". . 5. Given c= 90°, a- 109' 41'. Tirsults. C = 113' 28' 2", A = 142° 11' 38", 1} = 120° 15' 57". 6. Given c= 90', A = 131° 30', B = 120° 32'. Results. C = 1099 40' 20", a = 127° 17' 51", b=113' 49' 31". Given a = 76° 35' 36", b = 50° 10' 30", c= 40° 0' 10". Results. A = 121° 36' 20", B = 42° 15' 13", C-3415' 3". 8. Given A = 129º 5' 28", B 142" 12' 42", C = 105° 8' 10". Results. a = 135° 49' 20", b = 144° 37' 15", c=.60. 4'54". CAXBRIDGE : PRINTED AT THE UNIVERSITY PRESS.

·EDUCATIONAL MATHEMATICAL WORKS BY I. TODHUNTER, M.A., F.R.S. Euclid for Colleges and Schools. New Edition. 18mo, cloth. 38. 6d. Mensuration for Beginners. With numerous Ex-" amples. 18mo. cloth. Algebra for Beginner's. With numerous Examples. 28. 6d. New Edition. 18ino. cloth. 23. 6d. Key to the Algebra for Beginners. Second Edition. Crown Svo, cloth. 68. 60. Trigonometry for Beginners. With numerous Examples. Third Edition. 181o. cloth. 28. 6d. Mechanics for Beginners. With numerous Examples. Secoud,Edition. 18mo. cloth. 43. 6d. Algebra for the use of Colleges and Schools. With numerous Examples. Fifth Edition. Crown 8vo. cloth. 78. 6d. Key to the Algebra for the use of Colleges and Schools. Crown 8vo. cloth. A Treatise on the Theory of Equations. Second Edition. Crown 8vo. cloth. 78. 6d. 108. 6d.

·Plane Trigonometry for Colleges and Schools. With nuinerous Examples. Fourth Edition. Crown 8vo. cloth. 58. A Treatise on Spherical Trigonometry for the use of Colleges and Schools. With numerous Examples. Third Edition.' Crown 8vo. cloth. 48. 6d. A Treatise on Conic Sections. With numerous Examplek, Fourth Edition. Crown 8vo. cloth. 78. 6d. With A Treatise on the Differential Calculus. numerous Examples. Fifth Edition. Crown 8vo. cloth. 108.6d. A Treatise on the Integral Calculus. With numerous Examples.' Third Edition. Crawn 8vo. cloth. Ios. 6d. Examples of Analytical Geometry of Three Dimensions. Second Edition. Crown 8vo. cloth. 48. 15 A Treatise on Analytical Statics. With numerous Examples. Third Edition. Crown 8vo. cloth. 103. 60. A History of the Mathematical Theory of Probability, from the time of Pascal to that of Laplace. 8vo. cloth. 188. Edited by Me TODHUNTER. An Elementary Treatise on Differential Equations. By GEORGE BOOLE, D.C.L., F.R.S. A New Edition, edited by I. TODHUNTER, M.A., F.R.S. Crown 8vo. cloth. 148. A Supplementary Volume, Crown 8vo. cloth. 8s.6d. MACMILLAN AND CO. LONDON AND NEW YORK.

·No text on this page.

·No text on this page.

·ID93