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·SPHERICAL TRIGONOMETRY

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·CALCUTA PAINOVIT
ROVANCEMENTO
TEXT BOOK
OF
SPHERICAL TRIGONOMETRY
BY
PRAMATHA NATH MITRA, M.A.
LECTURER IN PURE MATHEMATICS IN THE UNIVERSITY OF CALCUTTA
SOMETIME SECRETARY OF THE CALCUTTA MATHEMATICAL SOCIETY
PUBLISHED BY THE
UNIVERSITY OF CALCUTTA
1935

·PRINTED AND PUBLISHED BY BHUPENDRALAL BANERJI
AT TR QALOOTIA UNIVRRSITY PRESS, SENATE TROBA. CALQONIA
Rog. No. 815B, Nov., 1985-500

·Το
THE LOVING MEMORY
OF
MY REVERED FATHER
BARODA PRASANNA MITRA

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·PREFACE The present work has evolved out of the lectures delivered to the Post-Graduate students of the University of Calcutta. It is intended as an introductory text-book on Spherical Trigonometry and an attempt has been made to present the subject matter in as simple a manger as possible. The book has been brought to the standard required for the examinations of Indiar Universities. It contains all the propositions which a student has and ought to learn to have a fairly comprehensive knowledge of the Trigonometry of Spheres, and thus it paves the way for higher study in Spherical Astronomy. As the book consists mainly of formulae and the applications thereof, a large number of examples has been appended for solution by the students. A short historical introduction has been given at the beginning, showing the successive stages of the development of the subject. It arose out of the growing need for the study of the heavens. It is interesting to note that the fundamental formulae were all known to Hindu Astronomers thousands of years ago and are of Indian origin, but owing to their Conservative spirit, any record of their work is wholly, 'wanting. It was Sürya Siddhanta which brought to light the achievement of Indian mathematicians, ·viii PREFACE on and this was followed by several works the subject, showing thereby that the ancient Hindus were far advanced in Astronomy. In the body of the book reference to authors of the respective theorems has in most cases been given. In the preparation of this book I had to consult the existing treatises and several memoirs on the subject, and my thanks are due to their respective authors. For the history of the subject, among other works, I was greatly influenced by the monumental · works of Dr. D. E. Smith and the late Dr. F. Cajori and my thanks are due to them. I am also indebted to Dr. S. M. Ganguli, D.Sc., P.R.S., Lecturer in Higher Geometry in the University of Calcutta, for his valuable suggestions. I have also to express my thanks to the authorities of the University of Calcutta for their consent to publish the book, and to the officers and the staff of the University Press, for the pains they have taken in the printing of the book. In conclusion, I hope that the present book will tend a little towards the advancement of Mathe. matical learning of our students ; it is for them that the book has been written and it is in their profit that I shall look for my reward. UNIVERSITY OF CALCUTTA : P. N. MITRA. July, 1935.

·CONTENTS ARTICLE PAGE Historical Introduction XV ... CHAPTER I SPHERE ... 1 2 ... 3 3 ... ... ... 4 5 1.1 Sphere 1.2 Intersection of sphere by a plane 1.3 Great circle and small circle 1.4 Axis and Poles 1.5-1.6 Theorems 1.7 Angle between two circles 1.8 Primary and secondary circles 1.9.1.13 Theorems 1.14 Arcs of small and great circles compared. Example worked out Examples 6 7 9 ... 11 12 ... CHAPTER II SPHERICAL TRIANGLE 1 13 2.1 2.2 2.8 2.4 Spherioal triangle Restriction of sides and angles Formation of & spherical triangle ... 13 15 ... Polar triangle 16 ·х CONTENTS ARTICLE PAGE 2.5-2.6 Theorems-properties of polar triangle 16. 2.7-2.9 Theorems on sums of the sides and 18 the angles. 2.10 Theorem-limit of the third angle 20 Examples 21 2.11 Lune 21 2.12 Antipodal triangles 23 2.13 Equality of triangles 24 2.14-2.15 Theorems 24 Examples 26. ... ... ... ... ... ... CHAPTER III RELATIONS BETWEEN THE SIDES AND ANGLES OF A SPHERICAL TRIANGLE 3.1 Fundamental Formulae-cosine Theo- 27 rem. ... 3.2 3.3 3.4-3.5 3.6 29 31 33 35. 3.7 Generalisation Expression for the sine of an angle Rule of sipes Analogous formulae in Plane Trigonometry. Distance between any two places on earth's surface Examples worked out Examples Formulae for half angles Formulae for sine of an angle 36 88 ... 3.8 3.9 42 44 46 ·CONTENTS xi 47 ... ... ... ... ARTICLE PAGE 3.10 Analogous formulae in Plane Trigonometry. Examples . 48 3.11 Formulae for cosine of sides 49 3.12 Analogous formula for plane triangle 50 8.13 Formulae for half sides 50 3.14 Formulae for sine of sides 52 Examples worked out 53 Examples 55 3.15 Cotangent formulae 57 3.16 Napier's analogies 58 3.17 Delambre's analogies 59 3.18-3.19 Napier's and Delambre's analogies 60 deduced. Examples worked out 64 Examples 68 ... CHAPTER IV RIGHT-ANGLED TRIANGLES ... 4.1 Formulae for right-angled triangle 4.2 Some important properties 4.3 Napier's Rules of circular parts 4.4 Quadrantal triangle 4.5 Trirectangular triangle 4.6 Direction angles and Direction cosines of a point. 4.7.4.9., Theorems on Direction cosines 73 76 76 78 78 79 80 ...

·xii CONTENTS ARTICLE PAGв 4.10 ... 83 84 87 90 90 ... ... .. 4.11 4.12 4.13 4.14 4.15 4.16 4.17 ... Direction cosines of the pole of an arc Examples worked out Examples Solution of right-angled triangles Two sides given-Example Two angles given - Example Hypotenuse and one side given Hypotenuse and one angle given One side and the adjacent angle given One side and the opposite angle given Example Application of Napier's analogies Example Solution of Oblique-angled triangles Examples 91 92 92 93 93 95 95 95 96 96 4.18 4.19 CHAPTER V PROPERTIES OF SPHERICAL TRIANGLES ... 5.1 Theorem on relation between three 98 arcs. 5.2 Particular cases 100 5.3 Spherical perpendiculars 101 5.4 Theorem 102 5.5-5.7 • Theorems on concurrency of three 103 arcs. 5.8 Theorem--concyclic points, spherical 106 transversal. ...

·CONTENTS xiii ARTICLE PAGE Theorem 5.9 5.10 5.11 108 Casey's Theorem 109 Normal co-ordinates of a point 110 Example worked out 111 Normal co ordinates w. r. t. trirect- 112 angular triangle. Examples 112 ... 5.12 CHAPTER VI AREA OF SPHERICAL TRIANGLE. SPHERICAL Excess 6.1 ... 6.2 6.4 6.5 6.6 0.7 Girard's Theorem-Area of spherical 115 triangle. Area of a polygon 117 Cagnoli's theorem-expression for 118 sin E. Expressions for cos }E and tan E 118 Formulae for colunar triangles 120 L'Huilier's theorem-expression for 121 tan E. The Lhuilierian 122 Expressions for sin F and cos JE 123 Exp. for sin (2A-E) and cos }(2A-E) 1(2A-E) 124 Geometrical representation of spherical 125 excess. Examples worked out 126 Examples 128 ... ... 6.8 6.9 6.10 6.11 . ...

·xiv CONTENTS ARTICLE PAGE APPROXIMATE FORMULAE -6.12 6.13 6.14 Legendre's theorem 130 Areas of spherical and plane triangles 132 Approximate value of spherical excess 133 Examples 134 CHAPTER VII CIRCLES CONNECTED WITH A GIVEN TRIANGLE ... ... 7.1 Inscribed and circumscribed circles 135 7.2 Radius of the incircle 136 7.3-7.4 Radii of the escribed circles 139 7.5 Radius of the circumcircle 140 7.6 Radii of the circumcircles of the 143 colunar triangles. 7.7 Inscribed and circumscribed circles 144 of the polar triangle. Examples worked out 145 7.8 Hart's circle ... 148 7.9 Spherical radius of Hart's circle 149 7.10-7.11 Angular distance of the pole of Hart's 151 circle from the vertices of the triangle, 7.12 Baltzer's Theorem 154 7.13 Theorem-intersection of circle by. 155 arc through a fixed point. 7.14 Lexell's locus Examples ... 158 ... ... ... 157 . ... ... .

·HISTORICAL INTRODUCTION The early history of the Science of Spherical Trigonometry is veiled in obscurity. In the prehistoric age, primitive men attributed every physical phenomenon to the agency of some Superhuman Being. To them the secret of the stars was closely connected with the secrets of human destiny. It was this that led the Hindus of India and the. Babylonian shepherd to observe the stars and to speculate on their meaning. Thus developed the folklore in India as also in the temples along the Nile and in Mesopotamią. As years ædvanced, observations of the heavens increased, which led to the measurement of angles, and thus the science of Astronomy had its beginning. The ancient Hindus however left no authentic record of their mathematical achievement. They were very conservative and would hardly impart their knowledge to their friends and disciples. Moreover they had little sympathy with those outside their own caste. It is only in some special case that a favourite disciple could acquire the knowledge and learning of his teacher. With the passing away, of a master mind, all his mathematical achievements were lost in oblivion. There is sufficient evidence to show that schools existed very early in India, where mathematics was looked upon as a very important ·xvi HISTORICAL INTRODUCTION branch of learning, but for the reasons aforesaid a general literature on the subject is wholly lacking. All that we can learn of them are gathered from the two great epics, the Mahabharata and the Ramayana, the Vedas and other ancient literatures which show that the Hindus from ancient times paid considerable attention to astronomy.* The oldest astronomical instrument dates as early as 1800 B.C. The study of scientific astronomy began in Greece with Thales (640-546 B.C.). He succeeded in dicting a solar eclipse which occurred on the 28th May, 585 B.C. Pythagoras (580-500 B.C.) asserted that Earth was spherical in shape. His teachings reveal much more of Indian than of the Greek civilisation in which he was born. It was left for Parmenides of Elea (460 B.C.) to teach at Athens the doctrine of the sphericity of the Earth. Eudoxas of Cindus (408-355 B.C.) is said to have introduced the study of spherics (mathematical astronomy) in Greece, Euclid of Alexandria (fl. 300 B.C.) wrote a book called Phaenomena dealing with the celestial sphere. Eratosthenes of Alexandrix (274--194 B.C.) took the noteworthy step in geodesy by his measurement of the circumference and diameter of the Earth. He also found the obliquity of the ecliptic * G.«Oppert, On the Original Inhabitants of Bharatavarsa or India, London, 1893. R. C. Dutt, A History of Civilisation in Ancient India, London, 1893.

·8 HISTORICAL INTRODUCTION vii to be 23°51' 20". Archimedes of Syracuse (287-212 B.C.) devoted a portion of his work on sphere. As yet ve have got nothing which can be called trigonometrical. Hipparchus of Nicæa (180-125 B.C.) wrote famous work on astronomy, in which he needed to measure angles and distances on a sphere, and bence he developed a kind of Spherical Trigonometry. He also worked out a table of chords, i. e., of double sines of half the angle, and thus was begun the science of Trigonometry. Menelaus of Alexandria (A. 100 A.D.) wrote a treatise on sphere Sphaericorum Libri III dealing with geometrical properties of spherical triangles. His propositior. Regula sex quantitatum is well known. He also wrote six books on the calculation of chords. The interest in astronomy had induced more progress in spherical rather than in plane trigonometry. Claudius Ptolemaeus (85-165 A.D.) brought together in his great work, Almagest in 13 books, the discoveries of his predecessors. He dovoted chapters of his first book to trigonometry and spherical trigonometry. He elaborated the table of sines already used by Hipparchus. He created, for astronomical use, & trigonometry remarkably perfect in form. Pappus of Alexandria (f. 300 A.D.) devoted his sixth book in Mathematical Collections to the treatment of sphere. From 2000 B.C. down to 300 B.C. we have no record of Indian astronomy save the glimpses we ·xviij HISTORICAL INTRODUCTION bave from the Vedic writings. The Vedic literatures were probably written about 2500-1500 B.C., though composed much earlier ; the Vedangas were written several centuries later. The ritualistic rules of the Sulvasutras were composed about 500 B.C. The Hindus were in the habit of putting into verse all mathematical results they obtained, and of clothing them in obscure and mystic language, which though well adapted to aid the memory of him who already understood the subject, was often unintelligible to the uninitiated. From the period of invasion of India by Alexander the Great in 327 B.C., there was regular intercourse bet veen the Hindu and Greek mathematicians, which influenced their respective astronomies to a certain extent. Before the beginning of the Christian era, there were numerous invasions from the North which seriously interfered with the spread of Greek science, and in the fourth century, A.D., with the appearance of Surya Siddhanta-the first important work on Astronomy in India-we find the astronomy of Greece replaced by the Astronomy of Hindus. The mathematical formulao of Sulva. sutras now gave place to the mathematics of stars. Spherical trigonometry and astronomy were treated scientifically by Aryabhatta (475-550 A.D.) in his Aryabhatiyam and Gola. Next comes Varahamihira * (505-587 A.D.) whose work Panca Siddhāntika shows * According to some tradition Yaraha and Mihira are, twodifferent persons-father and son.

·• HISTORICAL INTRODUCTION XIX an advanced state of mathematical astronomy. .He describes the five Siddhantas which had been written before his time but places the Sürya Siddhanta at the head. Among the five is the Paulisa Siddhanta which contains an excellent summary of early Hindu Trigonometry. Varahamihira taught the sphericity of the earth. The most prominent of the Hindu mathematicians, of the seventh century, was Brahmagupta, who was born in 598 A.D. He wrote his astronomical works Brāhma-sphuta-siddhanta in 628 A.D. and Khandakhādyaka in 605 A.D. It was he who taught the Arabs astronomy long before they became acquainted with Ptolemy's work. The famous Sindhind and Alarkand of the Arabs are the translations of the two books of Brahmagupta. The cosine and sine theorems for oblique-angled spherical triangles are implied in the rules of Varahamibira and Brahmagupta. The triadic relations for right-angled spherical triangles were known to the Hindu mathematicians and were used by them to solve spherical triangles. In the reign of Caliph Almansur of Bagdad a Hindu Astronomer named Kankah went to his court with astronomical tables * in 766 A. D., which were translated into Arabic. Thus Hindu mathematics * It is generally believed that this was the Brāhma-sphutasiddhānta of Brahmagupta, and the name Sindhind is derived from the word Siddhanta. A Persian named Yaqub ibn Tariq also went to the court of the Caliphs about this time and probably assisted in translating the works of Brahmagupta.

·XX HISTORICAL INTRODUCTION, and astronomy came to be known to the scholars at Ragdad. This was known as Sindhond and contained the important Hindu table of sides. After this time to the year 1000 A.D. very little progress was made in India. Mahavira (Al. 850 A.D.) seems to have made efforts to improve upon the works of Brahmagupta. In the meantime the knowledge of India passed into the keeping of Arabs. The chief Arab writer en astronomy was Albategnius (A. 920 A.D.). Like the Hindus ho used half chords instead of chords. Some mathematicians are of opinion that he discovered the coside formula, but there is no evidence to sbow that he had any real knowledge of spherical trigonometry. In fact, he borrowed it from the Hivdu astronomy. Abu'l Wefa (940—998 A.D.) and his contemporary Abu Nasr tried to systematise the older knowledge but it was the Persian astronomer, Nasir ed-din al-Tusi (1201_1274 A.D.), whose work Shakl al-qattâ reveals trigonomtry as a science by itself. Among the Hindu writers from 1000-1500 A.D., the first was Sridhara who was born in 991 A.D. but he did not contribute much to the science of Spherical Trigonometry. The other writer of prominence is Bhaskara (1114 1185 A.D.), whose Siddhanta Siromani contains a book, Goladhia, devoted to astronomy and sphericity of the earth. He gave a method of constructing a table of sines for every degree. With the decline of Bagdad, the study of spherical triangles, for astronomical work, assumed greater importance in ·HISTORICAL INTRODUCTION xxi Spain. Gabir ben Allah of Sevilla (1140 A.D.) wrote on spherical trigonometry and introduced the "rule of four quantities." By the 14th century England came to know of the Hindu Trigonometry through the Arab Trigonometry. Among tho orodern writers to exhibit Trigonometry as a science, independent of Astronomy, was the German mathematician Jobann Müller, better known as Regiomontanus (1436–1476). His work De triangu. lis omnimodis Libri V, written in 1464, may be said to have laid the foundation for later works on plane and spherical trigonometry. Copernicus (1473–1543) com. pleted come of the works left unfinished by Regiomontanus, in his De Lateribus et Angulis Triangulorum (1542). The Danish astronomer Tycho Brahe also gave the cosine formula in 1590. With the French mathematician Vieta (1540—1603) began the first systematic development of the calculation of plane and spherical triangles. The theorem for cosine of angles was given by Vieta in 1593. The cotangent theorem was given in substance by him but was afterwards proved by Snellius in 1627. The name Trigonometry first appeared in an important work on Trigonometry by the German mathematician Pitisous (1561–1613) in 1595. Albert Girard (1595-1632) published at the Hague, in 1626, a noteworthy work on Trigonometry, in which be made use of the spherical excess, in finding the area of a spherical triangle. This also appeared in his Invention nouvelle en l'Algébre in 1629. The area of a spherical triangle was also given by Cavalieri ·XX11 HISTORICAL INTRODUCTION (1598-1647) in his Directorium generale (Bologna, 1632), and afterwards in hiş Trigonometria plana et spherica (Bologna, 1643). Napier (1550—1617) replaced the rules for spherical triangles by one clearly stated rule, the Napier's analogies, published in his Mirifici Logarithmorum canonis Descriptio in 1614. He also gave two rules of circular parts, which included in tiem all the formulae for right-angled spherical triangles. The properties of the polar triangles were discovered by Snellius (1591-1626 A.D.) in his Trigonometria, published posthumously at Leyden in 1627. Euler (1707--1783 A.D.) gave a fresh impetus to the study of the subject by publishing several memoirs in the Royal Academy of Berlin and in the Acta Petropolitana. Delambre published his analogies in 1809. Valuable contributions to the subject were also made by Lagrange (1736-1813), Lhuilier (1750-1840), Legendre (1752-1833). Gauss (1777–1855), Lexell (1782), Chasles (1831), Schulz (1833), Gudermann (1835), Borgnet (1847), Neuberg, Yon Staudt (1798. 1867) and Simon Newcomb (1835-1909), E. Study (1893) and F. Meyer. The case for spherical triangles with sides and angles not necessarily less than 7 is generally ascribed to Mobius * but it seems that Gauss + had not only thought of this generalisation, but had worked it out. See Gesellschaft der Wissenschaften zu Leipzig, 1860, p. 51. + See, Theoria motus Corporum Coelestium, 1809, $ 54.

·SPHERICAL TRIGONOMETRY CHAPTER I SPHERE 1.1. Sphere. A sphere is a solid figure such that every point of its surface is equally distant from a fixed point within it, which is called the Centre of the Sphere. Any straight line joining the centre of a sphere to any point on its surface is called a Radius, and the straight line drawn through the centre and terminated both ways by the sphere is called a Diamcter of the Sphere. Al B' 0 B Pi A sphere can be generated by the revolution of a circle round any of its diameter as axis.

·2 SPHERICAL TRIGONOMETRY 1.2. Intersection of Sphere by a plane. If a plane intersects a sphere, the resulting section will be some curve on the surface of the sphere and we prove below that The section of the surface of a sphere by a plane is a circle. X А. c B D Let ABX be the section of the sphere made by a plane and let O be the centre of the sphere. Draw OC perpendicular to the plane of ABX. Take any point D on the section ABX and join CD and OD. Now OCD is a right-angled triangle, for OC is perpendicular to the plans ABX and hence perpendicular to CD. Therefore CD2=OD2-002. But OD is constant being radius of the sphere and OC is constant for O and C are fixed points, and hence CD is of constant length. Thus any point D in the section ABX is equally distant from the fixed point C in its plane, that is, ABX is a circle of which is the centre, ·GREAT CIRCLE AND SMALL CIRCLE . 3 1.3. Great Cirole and Small Cirole.* When the plane intersecting the sphere passes through the centre of the sphere, its circular section is called a Great Circle, thus AB is a Great Circle. When the section does not pass through the centre it is called a Small Circle, thus A' B' is a Small Circle. (See figure of Art. 1.1.) The solid cut off by the plane of a great circle is called a Hemisphere, and that cut off by the plane of a small circle is called a Segment of the sphere. Note 1.-Only one great circle can be drawn through two given points on the surface of a sphere, for its plane must pass through the centre of the sphere, and three non-collinear points uniquely determine a plane. The great circle is unequally divided at the two points, and by the arc jjining the two points we shall always mean the smaller of the two. But if the two given points be the extremeties of a diameter, an infinite number of great circles can be drawn through them. (See figure of Art. 1.1.) Note 2.-The shortest arc that can be drawn on the surface of a sphere joining two poir on it, is the great circular arc through them, for the shortest arc must have the least curvature, and so it must belong to the circle of the greatest radius, i.e., the great .circle. 1.4. Axis and Poles. The Axis of a circle on a sphere is that diameter of the sphere which is perpendicular to the plane of the circle. The extremeties The nomenclature is perhaps due to the fact that the radius of a great circle (wbich is the same as the radius of the sphere) is always greater than that of any small circle, as is evident from the relation CD2=OD2_002 of Art. 1.2.

·4
SPHERICAL TRIGONOMETRY
of the axis are called the Poles r* of the circle. Thus.
if PP (fig. of Art. 1.5) is perpendicular to the small
circle AB, P and Pl are its poles, of which the nearer
pole P will usually be denoted as the pole. The poles
of the great circle are equidistant from the plane of
the great circle. Any point and the great circle of
which it is the pole are termed pole and polar with
respect to each other.
EXAMPLE
Shew that the line joining the centre of the sphere to
the pole of a small circle passes through its centre.
1.5. Theorem. The pole of a circle is cquidistant
from every point on the circumference of the circle.
P
С
B
0
P'
Let O be the centre of the sphere and AB any
circle on it of which C is the centre, and P and P'
are the poles. Take any point D on AB. Join CD
and PD. Then PD2= PC2+CD2 =constant.
*
The expression pole of a circle is due to Archimedes of
Syracuse (287-212 B.C.).

·ANGLE BETWEEN TWO CIRCLES Now as the chord Þd is constant, therefore the arc of the great circle intercepting PD is also constant for all positious of D on the circle AB. Thus the distance of the pole of a circle from every point on its circumference is constant whether the distance be measured by a straight line or by a great circular arc. The great circular arc PD joining the pole P of the circle AB to any point D on its circumference, is called the Spherical Radius of the circle AB. The spherical radius of a great circle is a quadrant. (See Art. 1.9.) 1.6. Theorem. Two great circles bisect each other. The plane of each great circle passes through the centre of the sphere. Hence the line of intersection of these planes is a diameter of sphere, as also of each great circle. Therefore the great circles are bisected at the points where they meet. 1.7. Angle between two circles. When two circles intersect, the angle between the tangents at either of their points of intersection is called the angle between the circles. If these circles are great circles, their planes pass through the centre, and their line of intersection is a diameter of the sphere to which the tangents are perpendicular and hence the angle between the tangents is the angle of intersection of the planes.* Thus When two planes intersect, the angle between them is measured by the angle between any two straight lines drawn one in each plane, at ary point on their line of intersection and perpendicular to it.

·6. SPHERICAL TRIGONOMETRY The angle of intersection of two great circles is equal to the inclination of their plancs. 1.8. Secondary Ciroles. Great circles which pass through the poles of another great circle are called Secondaries to that circle, which again is termed Primary circle in relation to them. Thus, in the figure, ABC is the primary circle and all the circles through P and pl are secondaries to it. It is evident that there can be an infinite number of such secondaries, the planes of which intersect in the line PP, the axis of the primary circle. lo А) B C P' Since PPl is perpendicular to the plane ABC, any plane passing through PP! is also perpendicular to the plane ABC. Hence Any great circle and its secondary cut each other at right angles.

·GREAT CIRCLE AND ITS SECONDARIES, 7 Again since PO is perpendicular to, OA and oc, AOC is the angle of inclination of the planes of PA and PC, and this is measured by the arc AC. Hence the angle between the circles PA and PC is measured by the aro AC, i.e., The angle between any two great circles is measured by the arc intercepted by them on the great circle to which they are secondaries. 1.9. Theorem. 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. (Fig. of Art. 1.8.) Let P be a pole of the great circle ABC and O the centre of the sphere. Join P0. Then PO is perpendicular to the plane ABC and hence perpendicular to OA, OB, OC and OD. Hence each of the angles POA, POB, POC and POD is a right angle, i.e., the arc PA, PB, PC or PD is a quadrant. 1.10. The Converse Theorem. If the arcs of groat circles joining a point on the surface of a sphere with two other points on it, which are not opposite extremities of a diameter, be each a quadrant, then the first point is a pole of the great circle passing through the other two. For if PA and PC (fig. of Art. 1.8) be each & quadrant, the angles POA and POC are right angles. Therefore PO is perpendicular to OA and OC, and

·8
SPHERICAL TRIGONOMETRY
hence perpendicular to the plane AOC, i.e., P is
a pole of the great circle AC.
1.11. Theorem. If two arcs of great circles,
which are not parts of the same great circle, be drawn
from a point on the surface of a sphere such that their
planes are at right angles to the plane of a given
circle, then that point is a pole of the given circle.
Since the planes of the two arcs are at right
angles to the plane of the given circle, their line of
intersection is also perpendicular to the plane of the
given circle; and as it passes through the centre of
the sphere, it is the axis of the given circ Hepce
the given point is a pole of the circle.
1.12. Theorem. The points of intersection of tuo
great circles are the poles of the great circle passing
through the polcs of the given circles.
Р
B
10.
M
А
N
Let the two great circles intersect at A and B;'
and let P and Q be their poles. Join PA. and QA.

·DISTANCE OF POLES OF TWO CIRCLES 9 Then PA and QA afe each & quadrant (Art. 1.9) and hence A is the pole of the great circle PQ (Art. 1.10). Similarly B is the other poie, 1.13. Theorem. The angle between two great circles is equal to the angular distance between their poles. For, taking the figure of the last article, A is the pole of the circle PQ; hence AM and AN are each quadrant. The angle between the circles AMB and ANB is measured by the arc MN (Art. 1.8). Also PM and QN are quadrants and the angular distance of the poles is measured by the arc PQ. Therefore a arc PQ=PM -QM=QN-QM=MN. arcs Since these are equal, they will subtend equal angles at the centre. Hence joining OP, OQ, OM and ON, we have angle POQ=angle MON, i.e., the angle subtended at the centre of the sphere by the arc of a great circle joining the poles of two great circles is equal to the inclination of their planes. * 1.14. To compare the arc of a small circle subtending any angle at its centre with the arc of a great circle subtending an equal angle at the centre of the sphere. * It is obvious that the angle between the two planes is equal to the angle between their perpendiculars OP and OQ.

·10, SPHERICAL TRIGONOMETRY Let ab be the arc of a small circle whose centre is C and whose' pole is P. Let O be the centre of the sphere. Then OP is at right angles to the plane aC). OP is also at right angles to the plane of the great circle of which P is a pole. Through P draw great circles PaA and PbB to meet this great circle at A and B. Then OP is perpendicular to OA, OB, Ca and Cb. Hence either of the angles aCb or AOB measures the angle between the planes POA and POB and therefore Zacb= LAOB. P b C a 0 B A arc ab arc AB Hence radius Ca radius 0-1 arc ab arc AB Ca ОА Cu Oa = sin POa=cos or a=cos Aộa. arc ab: Thus =arc AB cos Ada. i.e., Distance betwcen two places on the same parallel of latitude = Difference in their longitude multiplied by cosine of their common latitude.

·EXAMPLE 11 EXAMPLE WORKED OUT . On a sphere whose radius is a a small circle of spherical radits, e, is described, upd a great circle is described baving its pule on the small circle, show that the length of their common chord is 27 ✓ - cos 20 BID (Science and Art Exam. Papers.). Let O be the centre of the sphere and C the centre of the small circle. Then OC is perpendicular to the plane of the small circie. Take any point P on the small circle as the pole of the great circle. Then LPOC=the angulør radius of the small circle=8 and hence OC=r cos 0 and CP= sin , when is the radius of the sphere. Let c be the length of the common chord and d the length; of the perpendicular froni Coo it. Then 2 () ) = x2 sin? 8 – d Again since the angle between OC and the plane of the great. circle is 90°-0, we have d cot a Or d=rcos cot . rcos 0 2 Therefore =p2 gin2o -p2 cos? o cot? 8 2 1%(sin? o-cos26) Bin? 0 Or, 27 sin - cos 20 N.B.-For a real section 20 must be greater than 90° and hence the negative sign under ibe radical sign.

·12 SPHERICAL TRIGONOMETRY EXAMPLES 1. Shew that any great circle is the locus of the poles of all its secondaries. 2. Shew that the angle between the plane of any circle and the plane of a great circle which passes through its poles is a right angle. 3. Two equal small circles are drawn touching each other. Sbew that the angle between their planes is twice the complement of their spherical radius. (Science and Art Exam. Papers.) 4. The angle subtended at the centre of a circle by two points on it is equal to the angle sobtended by them at its pole. 5. If two great circles are equally inclined to a third, their poles are equidistant from the pole of the third. 6. If a point is equidistant from three great circles, it is also equidistant from their poles. 7. If two spheres intersect each other, shew that their curve of intersection is a circle.

·CHAPTER II SPHERICAL TRIANGLE 2.1. Spherical triangle. A spherical triangle is a triangle formed by three arcs of great circles on the surface of a sphere. The arcs are spoken of as the sides, and their angles of inclination at the points wbere they meet, the angles of the spherical triangle. As in plane trigonometry, the angles are usually.. denoted by the letters A, B, C and their opposite sides by the letters a, b and c. The angles and the sides are sometimes spoken of as clements or parts of a spherical triangle. Unless stated to the contrary, all arcs drawn on the surface of a sphere will be taken to be arcs of great circles. . 2.2. Restriction of the sides and the angles, Two points on the surface of a sphere may be taken to be joined by either of the two segments of the great circle passing through them. Hence we can have eight triangles having for their vertices A, B and C. So to avoid ambiguity and to simplify our study it has been conventional (as in Art. 1.3, note 1) to mean by any of its sides, the lesser segment of the great circle passing through the two corresponding vertices. Thus we get one triangle ABC each side of which is less than a semicircle, and we denote this particular triangle 88 the spherical

·14
SPHERICAL TRIGONOMETRY
triangle ABC. Thus in the figura, triangle ABC is
that one formed by the arcs ADC, AB and BC.
D
А
B
E
It follows from the above that each angle of a
spherical triangle must be less than two right angles.
For consider the triangle ABC having the angle
B greater than two right angles. Produce the arc
AB to meet the circle ACF at E. Then the arc AFE
is a semicircle and hence the arc AEC is greater than
a semicircle. Thus the triangle ABC having the
angle B greater than two right angles is formed by
the arcs AB, BC and AEC of which the latter is
greater than two right angles. Such a triangle we
have excluded from our consideration. Hence we
conclude that
The sides and the angles of a spherical triangle
must cash be less than two right angles.
The sides and the angles of a spherical triangle
will generally be expressed in circular measure.

·SPHERICAL TRIANGLE 15 2.3. Formation pl a Spherical triangle. Let O be the centre of the sphere and suppose three planes form & solid angle at 0. These planes intersect the surface of the sphere in arcs of great circles AB, BC and CA which form. the sides of the spherical triangle ABC. А 0 с B Now the place angle AOB = arc AB > radius OA arc BC arc AC 7 angle BOC= and angle AOC radius OB radius OC and as 04=OB=0C, we see that the arcs AB, BC and CA are proportional to the plane angles AOB, BOC and COA, which they subtend at the centre of the sphere. If Ab and Ac are tangents to the arcs AB and AC respectively, the angle A is equal to the angle bAc, which again is the angle between the planes AOB and AOC containing the sides AB and AC. Thus the angles of a spherical triangle arc the same as the inclination of the plane faces forring the solid angle at the centre o of the sphere

·16
SPHERICAL TRIGONOMETRY
2.4. Polar Triangle.* If a triangle is formed with
the poles of the sides of a given triangle as its angular
points, it is called a Polar Triangle with respect to
the given triangle. Thus if ABC be a given spherical
triangle and Al, B', C' be the poles of BC, CA and
AB, then A'B'C' is the polar triangle of ABC. The
triangle ABC is called the primitive triangle with
respect to A'B'C'. Since there are two poles for
each side, we should get eight such polar triangles
with respect to the given triangle. But we call that
particular one to be the polar traingle in which
the poles A', B', C' lie on the same side of their
polars as the opposite angles A, B, C.
2.5. Theorem. If onc triangle be the polar
triangle of another, then the latter will be the polar
triangle of the former,
BI
м.
C
В.
IN
с
Let ABC be a given triangle and A'B'C' be the
polar triangle. Join AB' and AC'.
* The properties of the polar triangle were discovered, by :
Snellius (1591-1626 A.D.). His Trigonometria was, published
(posthumously) at Leyden in 1627.

·PROPERTIES OF POLAR TRIANGLE . 17 Now since Bl is the pole of AC, the arc AB' is a quadrant, and since CH is the pole of AB, the arc AC' is also a quadrant. Hence 1 is a pole of B'C'. And since it and Allie on the sar:le siile of BC, Al' is less than a quadrant. Again as I is a pole of DI", and 1A is less than a quadrant, 1 and A' lie on the same side of B'C'. Similarly B is the pole of 1'0' and (" is the pole of 1'B', and B, B' lie on the same side of A'('', and (, (" on the same side of A'!!'. Therefore ABC is the polar triangle of A'B'C'. 2.6. Theorem. The sides and ungles of the polar trianyle are respectively ihe supplements of the angles and sides of the primitive triangle. are a Let M and N be the points of intersection of .11 and 1( by BIC" (see fig. of Art. 2.5). Then 1 М and IN each quadrant, because it is the pole of BIC"; ard the angle it is measured by the arc MN. Again B'X' and ('M are also quadrants. Hence B'N + C'M=B'C' + AN=2 right angles, Or BIC=7-A. Similarly AC=T-B and A'B'= -0. Again since ABC is the polar triangle of A'B'C', We have BC=-A', CA=7-B', and AB=-CI.

·18 SPHERICAL TRIGONOMETRY Hence denoting the sides of the triangle A'B'C' by al, b', c', we have al=h-A, bl=1-B, and ci=1-C. and Al=h-a, Bl=-b, and ('= -0. Note.-From the above property polar triangles are also termed Supplemental triangles. Any theorem involving the sides and angles of a spherical triangle necessarily holds good for the polar triangle also. Hence for any such theorem there is a supplemental theorem involving the opposite angles and sides, and it is obtained by changing the sides and angles of the original theorem into the supplements of the corresponding angles and sides respectively. 2.7. Theorem. Any two sides of a spherical triangle are together grcater than the third side. Let ABC be a spherical triangle and ( the centre of the sphere. Now any two of the three plane angles forming the solid angle at () is greater than the third. Thus LAOB + LBOC > LAOC or, AB OA + BC OA > 0.4 that is, the sum of the arcs AB and BC is greater than the arc AC. Cor. Any one side of a spherical polygon is less than the sum of all the others. EXAMPLE Shew that the difference of any two sides of a spherical triangle is less than the third side.

·SUMS OF THREE SIDES AND THREE ANGLES 19 2.8. Theorem. The sum of the three sides of , spherical triangle is less than the circamference of a great circle. Let ABC ho a spherical "riangle, and 0 the centre of the sphere. The sum of the plane angles AOB, BOC and COA forming the solid angle at O is less than 27. i.e., AB ОА + BC 04 CA ОА <25 or AB + BC + CA < 27.0A. Thus the sum of the sides is less than the circumference of a great circle. The angular measure of the sum of the sides is less than four right angles. Aliter. B A' c. Let the sides AB and AC be produced to meet at the point A'. Then the arcs ABA and ACA' are semicircles. Now any two sides of the triangle A'BC are together greater than the third. Hence we have A'B + AC > BC. Therefore, AB + A'B + AC + AC > AB + BC + AC. or, . ABA' + ACA > AB + BC + CA ·20 SPHERICAL TRIGONOMETRY i.e., the sum of the sides is less than the circumnference of a great circle. Note'.-The above proposition can be easily extended in the case of polygons. 2.9. Theorem. The sum of the Threc angles oj a spherical triangle is greater than two right angles and 108% than six right unyles. Let 13 be a spherical triangle. Since each of the angles 1, B and C is less than it, we have 1+ B+C < 37. Again u' +'+c' < 271, where a', b' and ' are the sides of the polar triangle of ABC'. But it'==-1, U=7--13, c'=7-C (Art. 2.6). Henre, - 1+-- B + - 2. or', A + B +- ( > T Thus I < 1 -- R--( <3.. 2.10. Theorem. The difference blur('11 any 1100 angles of u spherical triangle is less thun the supple. ment of the third angle. . Let ABC' be a spherical triangle and A'B'(" be its polar triangle. Now any two sides of A'B'C' are together greater than the third.

·LIMIT OF THE THIRD ANGLI 21 Hence, d'+b' > or, - A+,-B>HO IB <n+C Hence, '-( <1- -B and B-C <A-A. Similarly, A-B < -C. This theorem gives the limit of the third angle when two augles are given. EXAMPLES 1. Given two angles of a spherical triangle to be 145º and 30", find the limit of the third angle. Here A =145º and B=80° Hence 145° - 80° = 65° <-C or C < 180° -15°, i.e., less than 115°. 2. If the difference between any two angles be 90°, sbew that the remaining angle is less than 90°. 3. Shew that the difference of the oblique angies of a rightangled triangle is less than a right angle. 4. Shew that the sum of the apgles of a right-angled triangle is less than four right angles. . 2.11. Lune. I Lune is a portion of the surface of a sphere enclosed by two great semicircles. Thus

·22
SPHERICAL TRIGONOMETRY
in the figure, the semicircles ABA" and ACA" enclose
a lune. A" is the point diametrically opposite to A.
B
с
A"
The angle BAC is called the Angle of the Lunc.
The triangles ABC and A"BC are called Colunar
Triangles, because they together make up a lune.
The area of a lune can be easily expressed in
terms of its angle, for,
Area of Lune
Area of Sphere
Angle of Lune
277
or,
Area of Lune=4752 = 2Arº,
27
where r is the radius of the sphere and A the circular
measure of the angle BAC.
If B" and C" be points diametrically opposite to B
and C respectively, we get two other colunar triangles
of ABC, namely B'CA and C"AB.

·ANTIPODAL TRIANGLE
23
2.12. Antipodat triangles. If the points A", B"
and C" diametrically opposite to A; B and Cres.
pectively be taken to form a triangle, the triangle
A"B"C" is called th: Antipodal triangle to ABC.
А
CN
B
BV
A"
The arcs AB and A"B" join diametrically opposite
points. Hence they are parts of the same great circle
and are equal in length. So also the arcs AC and
A"C" are equal, as also the arcs BC and B"C". Again
the angle A is equal to the angle A" for they are
comprised by the great circles ABA"B" and ACA"C".
Similarly B=B" and C=C". Hence the triangles
ABC and A"B"C" have all their elements equal. If
the triangle A"B"C" be shifted from its place on the
surface of the sphere till B" falls on B and C" falls on
C, the point A" will not fall on A but will lie on the
opposite side of BC. That is the triangle A"B"C" is not
superposable, on triangle ABC. Such triangles are

·21 SPHERICAL TRIGODTRY called symmetrically cqual * as aistinguished fruin identically "qual or congruent triangles which are superposable on each other. 2.13. Two triangles op the same sphere are equal (symmetrically or identically) when they have the following elements of one triangle equal to the corresponding elements of the other triangle. or, (1) Two sides and the included angle, (2) Three sides, (3) Two angles and the adjacent side. (4) Three angles. or, ol, The cases (1) to (3) are analogous to plane geometry, but (4) has no such analogue. It is derived from (2) by the consideration of the supplemental triangles. 2.14. In this and the following articles are given some theorems of plans geometry which bold good in the case of Epherical triangles as well. One such case has already been dealt with in Art. 2.7. Theorem The angles at the base of an isosceles 8pherical triangle ure qual, and conversely if two angles of a spherical triangle are equal, the opposite sides are cqual * This term is due to Legendre (1752-1833). See his E'léments de Géométrie, Paris, VI, Def. 16, 1794.

·ISOSCELES TRIANGLE
25
arc.
Let ABC be 4, spherical triangle of which the
sides AB and AC are equal. Take D to be the
middle point of AC. Join :D by a great circular
Then the triangles ADB and ADC have their
corresponding sides equal, ench to each, and therefore
they are symmetrically equul. Hence the anglo
B=the angle C.
For the converse case, take the angle B=the angle
C, and let A'B'C' be the polar triangle of ABC.
Now b'=7-B and c'=A-0
and as B=C, we have b'=c; hence B'=C'.
Again b=r-B' and C=1-C'.
Therefore b=c, i.e., AB and AC are equal.
2.15. Theorem. If one angle of a spherical
triangle is greater than another, then the side opposite
to the greater angle is greater than the side opposite
to the less and conversely.
А
В.
с
D
Let ABC be a triangle of which the angle A is
greater than the angle B. Draw a great circular arc
2-II

·26
SPHERICAL TRIGONOMETRY
AD making the angle BAD = the angle ABD. Then
the arc AD=the arc BD.
But in the triangle ADC, AD+DC > AC.
Therefore BD + DC, 1.C., BC > AC,
The converse case is easily proved with the help
of the polar triangles.
EXAMPLES
1. When does a polar triangle coincide with the primitive
triangle ?
Ans. When each element equals ft.
2. If two small circles on a sphere touch each other, shew
that the great circle joining their poles passes through their point
of contact.
3. If a triangle is eguilateral, shew that its polar triangle is
also equilateral.
4. If two sides of a spherical triangle be quadrants, shew
that the angles at the base are right angles.
5. If all the sides of a spherical triangle be quadrants, all
of its angles are right angles.
6. If two sides of a triangle are supplerental, shew that the
opposite angles are also supplemental.
7. If two sides of a triangle are supplemental, shew that two
sides of its polar triangle are also supplemental.
8. The base of a spherical triangle is given : find the locus of
the vertex when the sum of the other two sides is equal to two
right angles.
Ans. A great circle having the middle point of the buse
as pole.

·CHAPTER III RELATIONS BETWEEN TEE TRIGONOMETRICAL FUNCTIONS OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 3.1. Fundamental Formulae. Erpression for the cosine of an angle in terms of the sines and cosines of the sides. А E 0 B D Let ABC be a spherical triangle and ( the centre of the sphere. At A draw the tangents AD and AE to the arcs AB and AC respectively. They lie in the planes AOB and AOC respectively. Let them meet OB and OC produced at the poinus D and E. Then the angle EAD is equal to the angle A of the spherical triangle. Join DE. From the triangle DOE, we have DEY=OD2 + OE22 OD.OE cos a.

·28 SPHERICAL TRIGONOMETRY Again from the triangle DAE, we have DE2=AD2 + AE2-2 AD.AE cos A. Hence by subtraction we have (=OD2-AD2 +0E2-AE2 + 2 AD.AE cos A -- 2 OD.OE cos a. = 2 OA2 +2 AD.AE cos A-2 OD.OE cos a for the angles OAD and OAE are right angles. ОА (А AE AD Therefore, cos a= + cos A, OE OD OE OD cos a=cos b cos c + sin b sin c cos A. i.c., Similarly, and COS b = =cos c cos a + sin c sin a cos B, cos c=cos a cos b + sin a sin b cos C. Hence COE CO8 a cos 6 cos c sin b sin c COS B: cos b - cos c cos a sin c sin a 3 COS C COS Cand cos a cos 6 sin a sin b The above relations are the fundamental formulæ * * The coside theorem was implied in the rules of ancient Hindu Mathematicians for finding the time-altitude and the alt-azimuth equations and the diurnal motion, and was used by them to solve spherical triangles. Cf. Panca Siddhantikā, IV, 42-44, aby Yarahamihira (505-587); Brühma Sphūta Siddhānta, III, 26-40; and Khandakhādyaka, III, 13, by Brahma Gupta born in 598 A.D.; and Suryasiddhānta, III, 34-35 (written about the 4th century). It was exhibited in a systematic formly

·COSINE THEOREM
29
of the spherical trigonometry. All other formul:
can be made to depend upon them.*
3.2. On referring to the figure of the last article
it is seen that the angles AOD and AOE are acute
angles, and hence the arcs b and c containing the
angie 4 are each less than a quadrant. No such
restriction, however, has been placed upon the arc a,
so that a may be greater than, equal to, or less than
a quadrant. We shall now show that the above
formule apply to all spherical triangles whether
the arcs be greater than, equal to or less than a
quadrant.
(1) Let one side b be greater than a quadrant.
с
C
B
Produce CA and CB to meet at
C.
Then
the German Mathematician Regiomontanus (1436-1476) in 1.160
ard afterwards by the Danish Astronomer Tycho Brahe about
1590. Euler also gave a proof of the theorem in bis Mémoires
de Berlin in 1753. Some are of opinion that it was discovered
by Albategnius (900 A.D.) who in fact borrowed it from the
Hindu Astronomy.
As was shown by Lagrange (1736-1813). See also Gauss
(1777-1855), Ges. Werke, Vol. IV, p. 401.

·30. SPHERICAL TRIGONOMETRY C'A=T-6, and C'B=r-a. Hence from the triangle ABC', we have cos BCI=cos AB cos AC! + sin AB sin ACI cos BAC', or cos a=cos b cos c + sin b sin c cos A. (2) Next let b and c be each greater than a quadrant. B А A' с Produce AB and AC to meet at A'. Then from the triangle A'BC, we have cos BC=cos AIC cos A'B + sin A'C sin A'B cos A', or cos a=cos b cos c+ sin l sin c cos A, for A'C=7—6, A'B=7-c and A= A'. (3) Thirdly let b be equal to a quadrant. From AB or AB produced cut off AD equal to it quadrant. Join CD.

·FORMULA FOR SINE OF AN ANGLE 31 Now if CD be a quadrant, C will be the pole of А В. D B AB, and the formula becomes 0=(). If CD be not a quadrant, we have from the triangle BCD, cos arcos BD cos CD + sin BD sin CD cos BDC = sin c cos A für COB BDC=0. The formula also reduces to tbis when h=4. (4) Lastly let b=c=. Then our formula reduces to cos arcos A, as is otherwise evident, since 4 is the pole of BC. Thus A=a. Thus our formula is universally true. 3.3. Expression for the sine of an angle. 1 We have COS A: COS a cos b oos o sin b sinc ·32 SPHERICAL TRIGONOMETRY Therefore gin? A=1- COB a cos b cos sin b sin c sinab sinac-(cos a cos b cos c)2 sinab sinac (1-cos2b)(1-cos2c) - (cos a sinab sinac cos b cos c)2 1-cos-a-cos? b) - cos2c +2 cos a cos.b cos c sinab sinac so that sin A= W{1-cos-a-cos2b-cos?c +2 cos a cosb cosc} sin b sinc As sin A, sin 6 and sino are all positive, the radical must be taken with the positive sign. For the sake of brevity and owing to the importance of the expression under the radical sign, we put 4n2=1-cos?a-cos2b-cosac +2 cos a cos b cos C, so that sin A= 2n sin b sin c sin BE 2n sin ( sin a > 21 and sin (= sin a sin b n is called the norm of the sides of the spherical triangle.* * This nomenclature is due to Professor Neuberg of Liege. Professor Von Staudt (1798 -1867) calls 2n the sine of the tria.egle ABC. See Crelle's Journal, XXIV, 1842, p. 252.

·RÚLE
RULE OF SINES
33
3.4.
From the value of sin A, we have at once
sin A
sin C
sin B
sin b
---
2n
sin a sin b sin c
sin a
ginc
1.6., the sines of the angles of a spherical triangle are
proportional to the sines of the opposite sides.
Owing to the importance of this result, we give
an independent proof of it in the next article.
3.3. Rule of Sines. The sines of the angles of a
spherical triangle are proportional to the sines of the
opposite sides.
с
2
X
B
Let ABC be a spherical triangle and 0 the centre
of the sphere. From A draw AX perpendicular to
the plane BOC, and AY and A2 perpendiculars on
OB and OC respectively. Join OX, XY and XZ.
3

·34 SPHERICAL TRIGONOMETRY Ther since AX is perpendiculai to the plane BOC, it is at right angles to OX, XY and XZ. Hence OAP=OX2 + AX2, AY2= AX2 + XY2 and AZ2 = AX2 + XZ2. Also OA2=OY2+ AY2=022 +AZ?. Therefore, OX=0A2-AX2=0Y2 + AYP-AX2 = OY2+ XY2 Similarly, OX? = 0A2-AX2 = OZ2 + AZ2 - AX2 =022 + XZ2. Thus XY and XZ are at right angles to OB and OC respectively. Now since AY and XY are in the planes OAB and OBC and are at right angles to their line of intersection OB at Y, the angle AYX measures the angle B of the spherical triangle. (Art. 2.3). Similarly angle AZX measures the angle C. Hence AX=AY sin AYX= AY sin B=0A sin c sin B and AX= AZ sin AZX=AZ sin C=0 A sin b siu C. sin c sin B Therefore sin o sin b Since B and C are any two angles of the spherical triangle, it follows that sin A sin B sin C sin b sin a sin c * This theorem appears in a different form in the 3rd book of the Sphaerica of Menelaus of Alexandria (100 A.D.). It was also known to Abd'l Wefâ (940-998) of Arabia and possible to liis contemporary Abd Nasr.

·FORMULAE IN PLANE TRIGONOMETRY 35 3.6. Analogous formulde in Plano Trigonometry. The sine and cosine formulæ in the previous articles bear some resemblance to the corresponding formulæ in Plane Trigonometry. In fact the latter war he derived from thu, former when the radius of the sphere is taken to be indefinitely great, for then the great circular arc reduces to a straight line ard the limiting form of the proposed formula becomes the formula for Plane Trigonometry. Let a, B, y be the lengths of the sides of the spherical triangle a B y ABC, then are the circular measures of the sides, T T ir being the radius of the sphere. From Art. 3.1 we have COS cos i cos b cusc sin o sin c a B γ COS COS COS T T T ain sin T T Expanding the sines and cosines in series, we get a? (1- + 72 27% + 272 2r2 COB A ....-(1 .)(1- (n.):-*...) 1 Hence, retaining terms involving only up to and taking ? COS A to be infinite, we have B? +g-a? 2By which is the expression for the cosine of an aogle in terms of the sides in Plane Trigonometry.

·36 SPHERICAL TRIGONOMETRY Similarly for the sine formula we Lave sin sin A sin B sin a sin sin B which on expansion becomes 1 as 31 13 T a + a(B?-a?) 6Br? + ..... B B 1 BS T 31 p3 + Hence taking r to be infinite, we have a sin A sin B 2 i.e., in a plane triangle the sides of the angles are proportional to the opposite sides.* any two places OD 3.7. Distance between Earth's surface. Let A and B be two places on Earth's surface and let their latitudes and longitudes beli, l, and de le respectively. Take P as the pole of the This formula is implied in Khandakhôdyaka, VI, 1, by Brahmagupta. See the English edition by ijP. C. Sen Gupta, p. 115,

·DISTANCE OF TWO PLACES
37
Equator and draw two secondaries to it through A
and B respectively meeting it at a and b*respectively.
А
B
2
b
Then Aa=l], Bb=l, and ab=1,-1.
Now from the triangle PAB, we have
cos AB=cos PA cos PB +sin PA sin PB cos APB
or denoting the arc AB by d, we have
cos d=sin l, sin ly+cos l, cos la cos (1,-12) ... (1)
This formula can be put in another form, from
which d can be obtained when A and B are very close
to each other. For we have
cos 8 = sin lq sin l,{cos2 (,-12)+sin? 1(11-12)}
+ cosl, cos l,{cos? 1(1,-12) - sin? (^,-1)}
1 = Cos (1-12) cosa 1(1,-1g) - cos (l,+lg)
sin? (1,-12).

·38 SPHERICAL TRIGONOMETRY Subtracting this from 1=cos2 1(4, -12) + sin? (^, -12) we get 8 sin 2 cos? 7(11-12) sin? 1.-12) + sinº 1947-12) cos2 {(1, +lo). (2) ... Hence when A and B are very close together, the approximate value of 8 is given by Se=(1-12)+(^, -12) cos? (?, +10). (3) .. EXAMPLES WORKED OUT Ex. 1. If D be any point in the side BC of a triangle ABC, shew that cos AD sin BC=cos AB sin CD +cos AC sin BD. We have cos ADB=C08 AB-COS AD COR BD sin AD sin BD and COB ADC=COB AC-Cos AD COS CD sin AD sin CD But cos ADB = -cos ADC. Hence CO8 AB sin CD + cos AC sin BD =cos AD(sin BD cos CD + cos BD sin CD) =COS AD sin BC. Ex. 2. In any triangle, shew that sip (A + B) cos a + cos 6 1 + COB C sin C ·EXAMPLES 39 We have sin (A + B) sin 'A cos B+cos A sin B sin C sin C CON CO6 a cos C + cos a-cos 6 cos c sinac sinac by Arts. 31 & 3.4 cos a + cos 6 1 + cos C Ex. 3. If a, B and y be the arcs joining the middle points of the sides of a spherical triangle A BC, sbew that COS a CO3 7 cos B b cos a с 1 + cos a + cos b + cos c a 6 C 4 COS cos COS 2 COS cos The arc a joins the middle points of band c. Hence we have с C COS a= COS cos + sin sin cos A 2 с = COS b 2 COS + sin sing cos a - cos b cos c sin b sia , by Art. 3.1 2 (1 + cos b)(1 + cos c) +008 - cos b cos c 4 cos COS lolos 1 + cos a + cos b + cosc b COB 2 C 4 cos COS a 1 + cos a + cos b + cos c a 4 COB b COB C cos COS 2 2 Similar expressions are obtained for cos B and cos y. result. Hence the ·40 SPHERICAL TRIGONOMETRY Ex. 4. In a spherical triangle ABC, great circular arcs a, B and yy are drawn from the vertices A, B and C perpendicular to the opposite sides and terminated by them. Shew that sin a sio a= sin b sin B = Bin C sin y :(1-cos-a-c032h-cosạc + 2 cos a cos b og c) (C.U., M.A. & M.Sc., 1932.) A F с ве D Let a, B and y meet the opposite sides in D, E and F respec. tively. Then from the triangle ABD, we have sin a = sin c gin B, by Art. (3.4). :. sin a sin a=sin a sin B sin c. Similarly from thỗ triangles BEC and BFC, we have sin B=sin a sin C and sin g=sin a sin B. Hence sin o sin a= =sin b sin B=sin C sin y=sin a sin b sin C, 2n Bin a sin 6 - 2n sin a sin b (1-cos?a-cos2b-cos?c + 2 cos a cos b cos c). Definition. The length of the great circular arc, drawn from the vertex of a spherical triangle perpendicular on the opposite side and terminated by it, is called an Altitude of the triangle. Thus in the above example a, B and g are the three altitudes of the triangle ABC.

·EXAMPLES
41
The above example shows that
The product of the sine of a sile and the sine of the cor.
responding altitude has the same nalue, whichever side be taken.
Ex. 5. Two ports are in the same parallel of latitude, their
common latitude being I and their differince of longitude 21; shew
that the saring of distance io sailing from one to the other on the
great circle, instead of sailing due east or vest is
2 r {* cos l-sin-(sin cos 2)},
^ being expressed in circular measura, and being the radius
of the earth.
(C. U., M. A. & M.Sc. 1931.)
Р
с
B
B
A
Let a and b be the two ports. Through the pole P of the
small circle ab draw great circular arcs PiA and PbB to meet
the great circle of which P is the pole (the equator) at A and B.
Then AB is the difference of longitude.
Let the arcual distance ab along the small circle be 8 and along
a great circle bed, so that their respective circular measures
d
wand
are
S
1
1

·42 SPHERICAL TRIGONOMETRY Now by Art. 1.14, arc ab arc AB =cos AO=cos 1. S 21 cos 1. 7 Again cos =cos ( -1) + sin?( -I) cos 24, 2 -sin?l + cos?l cos 24. d Hence 2 sin? =cos” 1-cosº I cos 21=2 cos 27 sin??. 27 .. sin d 21 =cos / sin , Or d =sin-1 (cos l sin ). 21 Hence the required saving of distance - 2r i cos l-2 r sinº' (sin a cos 1) = 27 {cos l-in-1 (sin a cos 7)}. EXAMPLES 1. If A=a, sbew that B and b are equal or supplemental, as also C and c. 2. The base BC of the triangle ABC is bisected at D. Show that (1) COS AB+cos AC == 2 cos AD, cos BD. (ii) sin BAD : sin C4D=sin b : sin c. 3. In an equilateral triangle, slew that (2) sec A=). + seca, a A 2 cos sin 2 2 1. (iii) tan2 a 2 1-2 cos A.

·EXAMPLES 43 4. If an angle of a triangle be equal or supplemental to the opposite side, sbew that 1-secla - sec%b-gecc + 2 sec a sec b sec c= 5. If 8 be the length of tbe arc joining the middle point of the side AB with the vertex C, sbew that COS cos a + cos b 2 cos c 6. The base BC of the triangle ABC is bisected at X, anil a point Y is taken on BC such that tbe LBAX=LCAY. Shew that sin BY : sin CY=sin? c: sin? b. 7, In a triangle ABC, a, b, y are drawn perpendiculars from the vertices A, B, C on the opposite sides. Sbew that sin a cos a=n W{cos2b + cosec- 2 cos a cos b cos c}, sin li cos B= V{cos?a + cosc - 2 cos a cos b cos c}, (iii) sin c cos g= ✓{cos?a + cosab- 2 cos a cos b cos c}. 8. Prove that 8n3 = sin’a sin?b sin'c sin A sin B sin C. 9. In any triangle, ebow that sin (A - B) sin C cos b-vos a 1-CUS C 10. If a' be the arc joining the middle points of the sides AB and A'C of the colunar triangle of ABC, shew that cos a 1 + cos a cos b-COS C 4 sin f bain fc ·44 SPHERICAL TRIGONOMETRY 11. If a, B and sy be the arcs joining the middle points of the sides of a spherical triangle, shew that when one of them is a quadrant, the other two are also quadrants. 12. A port is in latitude 1 (North) and longitude 1 (East). Shew that the longitudes of places on the Equator distant 8 from the port are cos 1 + cos cos/ (Science and Art Exam. Papers.) 13. Two places on the Earth's surface are distant, one e from the Pole and the other from the Equator, and their difference of longitude is 0; show that the angular distance between them is 008-'(sin 2 8 cos? 10). (Science and Art E.ram. Papers.) 3.8. Expressions for the sine, cosine and tangent of half an angle. We know that COS A=1-2 sina 2 1 Hence, sinº 4 cos A 2 COS a = 1 cos V cos C sin b sin c by Art. 3.1 cos (b --c) - cos a sin b sinc . sin f(a+b-c) sin f(a-b+c) sin b sinc ·FORMULAE FOR HALF ANGLES 45 Put 28=a+b+c, then 8 denotes the half of the sum of the sides of triangle, and a+b-c=2(8-c), a-b+c=218-6) so that sin2 A 2 sin (8-b) sin (s-c:) sin b sin c * Hence, sin ܠܐ |ܟܬ sin (8-b) sin (8-0) sin b sinc Similarly, sin B 2 v{ NI NE sin (s-c) sin (s-a) sin c sin a and sin o 2 sin (s-a) sin (s-6) sin a sin b Again, cog2 A 2 1+cos A 2 1+ COS a cok b cos c sin b sin c cOS a -> cos (b + c) sin b sin c sin : (a + bc) sin } (b +(-a) sin b sin c sin 8 sinn (8-a) sin b sinc Obtained by Euler (1707-1783) in 1753.

·46 SPHERICAL TRIGONOMETRY * A Hence COS VI sin s sin (s-a) sin b sin c 2 Similarly, cos B 2 NE sin s sin (s-b) sin c sin a and cos c 2 NE sin s sin (8-0) sin a sin b From the above results, we get * tan NE sin (s-b) sin (8-0) sin s sin (8-a) 2 tan B 2 NI sin (8 —csio (s - a) sin s sin (s-b) and tan c 2 NI ein ($-a) sin (s-b) sin s sin (8-0) The radicals in the results of this article must bo taken with positive sigos, since the half angles are each less than a right angle and hence their sines, cosines and tangents are all positive. 3.9. Again since sin 1=2 sin cos 2 we have sin A= 2 sin b sin c {sin s sin(s-a)sin(8 -- b)sin (8-c)}. * Euler, 1. c.

·FORMULAE FOR HALF ANGLES 47 Comparing it with the expression for sin A as given in Art. 3,3 we get n2=sin s.sin (s-a) sin (8-5) sin (8-0) 1 {1-cosa – cos2b -cosºc +2 cos a cos b cos c}.* 3.10. Analogous formulae in Plane Trigonometry. Taking a, b, a to be the lengths of the sides of the spherical triangle, we bave a, Baud as their circular measures. The T 1 r - a sin Asten siu } T T COS A 2 sin s sin (s-ai sin b sinc B sin sin ? T T where 2s' =a + B + y. Hence op expanding the sines and cosides, we have s + 1s 13 6 73 B r COS 4 s' -a 1 (s'-a)3 6 p3 7 1 73 p3 ] -- 1 83 6 3 r Thus retaining only up to the second power of r and taking , to be infinite, we get A COS VE s'(s' -a) βγ * Novi These expressions for n are given by Euler in Commentarii Petropolitana, Vol. IV, p. 158.

·48 SPHERICAL TRIGONOMETRY for the relation for a plane triangle. Similarly, sio А 2 (s'-B)(s'-) By and tan A = ✓ (s' -B)(s' - s'(s' -a) Again from the relation 2n 2{sin s sin (8 – a) sin (8-6) sin (8-c)}} sin A= sip b sin c sin b Bin C we get 2{s'(s' – a)(s' —B)(s' — x)}+ sin A By so that the area of the plane triangle ABC is 4 = {s'(&? –a)(s' —B)(8°—2)}}. This form is due to Heron of Alexandria (50 A.D.). EXAMPLES In any spherical triangle, shew that 1. tan 1A tan B= sin (8-c) sin s 2. cotA: cot fB : cot 1C=sio(s-a) : sin(s -b): sio (s-c). 3. sin 8= cos IB cos IC sin A sin a. gina, 4. sin (8-2)= sin B sin JC sin ja 5. sin 8 sin a sio b sin c sin } A sin JB sin fC=na, ginc 6. совес А cos JB сов 1С sin (8-6) ·FORMULA FOR COSINE OF A SIDE 49 3.11. Expression for the cosine of a side in terms of sines and cosines of the angles. Leu a', b' and c' be the sides and A', B' and C the angles of the polar triargle of ABC. Then by, Art. 3.1 we have cos a'=cos bl cos cl + sin bi sin d' cos A'. Substituting the values of a', b', cand A' from Art. 2.6 we have cos (1 - A)=cos (- B) cos (T-C) +sin (T-B) sio (--C) cos (-a), that is, COS AC cos B cos C+sin B sin C cos a. Similarly, cos B=-cos C cos A + sin C sin A cos b, and COS Ccos A cos B + sin A sin B cos c. These * can also be written as cos a cos A +cos B cos C sin B sin c. 1 cos b= cos B + cos C cos A sin C sin A and COS C cos C +cos A cos B sin A sin B * Tbese formulae are due to the French Mathematician Vieta (1540-1603) who published them in the eighth book of his Variorum de rebus mathematicis responsorum in 1595.

·50 SPHERICAL TRIGONOMETRY 3.12. Analogous formula for plane tringle. When, the radius of the sphere is taken to be infinite, we have a cos a=COS=1. T Hence the formula COS A -Cos B cos C + sin B sin C cos a becomes COS A = - cos B cos C + ein B sin C 80 that B+C=T-A Or A + B + C=, showing tbat the three angles of a plane triangle are together equal to two right angles. 3.13. Expressions for the sine, cosine and tangent of half of a side in terms of sines and cosines of the angles. a 1 cos a We have sin ? 2 2 = 1 cos A-+cos B cos C sin B sin c = = 1 cos A+ cos (B+C) sin B sin c cos 1 (A+B+C) COR 1 (B+C- A) sin B sin C ·FORMULAE FOR HALF SIDES 51 Putting 28 = A +B+C, we have a sin = 1'{- cos S cos (S - A) gin B sin C CA with similar expressions for sin b aad sin c. a 1 + cos a Again cus2 2 =}{ 31+ COS A +cos B cos C sin B sin c il non cos A+cos (B-C) sin B sin c cos }(A-B+C) cos }(A+B-C) sin B sin c cos (S-B) cos (S-C) sin B sin c a Hence COS N{co cos (South A S –C).}. Also tan = { cos S cos (S-A) cos (S-B) cos (8-C) o } 2 The radicals must be taken with positive signs since Ja is less than a right angle.

·52 SPHERICAL TRIGONOMETRY It is to be noted here that the value of S lies be. tween 1 and gir. Hence the value of cos S is negative and the values of S-A, S-B and S-Clie between -n and 17 (Arts. 2.9 and 2.10) so that their cosines are positive. Hence the expressions within brackets are positive so that the values of sin fa, cos ja and tan ja are all real and positive. The above results could have been obtained from the results of Arts. 3.1 and 3.8 by changing the sides and angles into the supplements of angles and sides. They illustrate the proposition that if a theorem holds good between the sides and angles of a spherical triangle, the theorem will remain true when the sides and angles are changed into the supplements of the corresponding angles and sides respectively. (Art. 2.6, note.) 3.14. Expression for the sine of a side. We have sin a -2 sin la cos fu 2 sin B sin C -cos S cos(S-1) cos(S-B)cos(S-C) --)! We shall use the symbol N to denote {-cos S cos (S-A) cos (S-B) cos (S-C)}}; 2N sin B sin c then sin a= 2N Similarly, sin b= sin C sin A and sin c= 2N sin A sin B ·EXAMPLES 53 Thus 2N=sin a sin B sin (=sin A sin b sin C =siu A sin B sin c. N is called the Norm of the angles * of the spherical triangle. EXAMPLES WORKED OUT Ex. 1. In any triangle shew that cos A + cos B 1-C08 C sio (a + b) sinc We have COB A = = -cos B cos C+ sin B sin C cosa and cos B= -cos A cos C + sin A sin C cos b. Adding these we get COS A +cos B - cos C (cos A +cos B) + sin C (sin B cos a + sin A cos b), whence, (cos A + cos B) (1 + cos C) (sin b cos a +gin a cos b) sinC. sin c by Art. 3.4. cos A + cos B sin (a + b) Thus 1 - cos C ginc ** Due to Professor Neuberg. Professor Von Staudt calls 2N sine of the polar triangle. Various expressions for N were given by Lexell in Acta Petropolitana, 1782, p. 49.

·54 SPHERICAL TRIGONOMETRY Ex. 2. If 8 and e' denote the angles which the internal and external bisectors of the angle ( make with the side AB, shew that COS cos A ~ cos B 2 cos fc and COS O' cos A +cos B 2 sin fc Let 8 and 8 be the lengths of the internal and external bisectors of the angle C and let them meet AB at D and I respectively, making with it the angles 8 and e'. Then from the triangle ACD, we have by Art. 3.11 cos A +cos e cos C sin o sin C COB Similarly from the triangle BCD, we have COS 8 cos B-Cos 6 cos C sin e sin c Equating these two values of cos d, we have COS A COB 8 ~ cos B 2 cos 0 Again from the triangles AEC and BEC, we have COB 8 cos A + cos d' cos (T + C) sin e' sin }(+ C) cos (1-B) + cos o' cos f(T-C), sin a' sin (1-() whence, cos Ꮎ . cos A +cos B 2 sin C ·EXAMPLES 55 1 EXAMPLES 1. If the side BC of the triangle ABC be a quadrant, ghew that cos A + cos R cos C=r 2. In any triangle, shew that cos A -cos B Bin (b-a) 1 + cos © sin c 3. In any triangle, shew that cos A + cos B 1 -- cos C sin (a-0) sir. C=(), and cos A - cos B sin (a + b) sin c=0. 1 + cos ( 4. In an equilateral triangle, shew that tun? o/ -1-2 cos A. 5. Shew that 4 N2=1-cos? A-cos? B-cos2 C-2 cos Acc8 B cos C. 6. Jf a, B, y be the arcs of great circlas dra vn from A, B, C perpendicular on the opposite sides and terminated br them, shew that (0) Bin A sin a= = sin B sip B=sin C sin y=2N, (ii) sin a sin a=sin b sin B=sin c sin g=2n. 7. Prove that sin a ginc n sin b sin B sin A sin C N 8. Prove that 2n2 N= sin a sin b sin c' and 2N2 sin A sin B sin c ·56 SPHERICAL TRIGONOMETRY 9. Sbew that * 2N (sin a sin b sin c sin? A sin? B gin? c). 10. Shew that 4nN sin a sin b sin c sin sip B sin ('. 11. Shew that C cogs tan he tan 2 cos (S- A) 12. Shew that a tan : tan cos(S-A): cos(S-B) : cos(S-C). 2 13. Shew that sin? a + sin? b + sinac ain? A + sin2 B + sina C 1 - cos a cos li cos c 1 + cos A cos B cos C (Dublin University Exani ination Papers.) 14. Shew that cos (S-A sin A cos tb cos fc. cos ja 15. Shew that - cos S sin ja sio fb sin cos dc n cos fa cos tb cos fc 16. Shew that cot ja cos (S--A)=cot fb cos (S-B) = cot dc cos (S-C) n -cot fa cot fb cot tc cos S 2 sin ju sin jo sip fc

·COTANGENT FORMULAE
57
3.15. Relations existing between two sides, the
inoluded angle and another angle.
Cotangent formulae.
I tary spherical triangle,
cot a sin b = cot A sin C + cos b cos C.
We have
cos a = cos b cos ( + sin b sin c cos A
== cos b (cos a cos b + sin a sin b cos C)
sin b sin a sin C cos A
sin A
+
by substituting the values of sin c and cos c.
Thus
cos a (1 - cos2 b)= sin a sin b cos b cos C,
+ sin a sin b sin C cot A,
or, cos a sin2 b==sin a sin b (cos b cos C+cot A sin C),
i.e.,
cot a sin b = cot A sin C + cos b cos C.
By proceeding similarly we
formulae, namely,
can get five other
cot b sin a = cot B sin C + cos a cos C.
cot b sin c = cot B sin A + cos c cos A.
cot c sin b = cot C sin A + cos h cos A.
cot c sin a = cot C sin B + cos a cos B.
cot a sin c = cot A sin B + cos c cos B.

·58 SPHERICAL TRIGONOMETRY Of the four elements entering into any one of the formulae it will be noticed that one side lies between two angles and one angle is included by the two sides, and if we denote them by 1 and 2, and the remaining side and angle by 3 and 4 respectively, all the formulae* are expressed in the form sin 1 sin 2 cos 1 cos 2 = cot 4 cot 3 3.16. Napier's analogies. We have tan 1(A + B) tan fC= tan A tan 1C +tan B tan C 1 -tan 4A tan B Substituting the values of tangents from Art. 3.8 we get sin(s-a) + sin(8-5) tan 1(A + B) tan C= sin sa sin(s—c) cos (a - b) cos (a + b) ... (1) Thus, tan 3(A+B) tan C=cos (a - b) cos (a + b) Similarly, sina-b) tan (A - B) tan C= sin(a+b) (2) Dr. Leathem states the formula in the form (cosine of inner side) (cosine of inner angle) (sine of inner side) (cotangent of other side) - (sine of inner angle) (cotangent of other angle).

·ANALOGIES OF NAPIER AND DELAMBRE 59. Again by substituting the elements of the polar triangle in (1) and (2), or proceeding as in (1) and (2) with tangents of half sides (Art. 3.13) we get tan /(a + b)_cos 1(n-B). cos (A + B) (3) tan 10 tan to tan fla-b)_sin }(A-B). and (4) sin }(A + B) The above four formulae are known as Napier's analogieg.* As a, b and C are less than 1 (Art. 2.2), cos i (a - b). and tan 10 are essentially positive. Hence in (1) tan 3(A+B) and cos f(a+b) must have the same sign. Therefore }(A + B) and f(a+b) must either be both greater than it or both less than 1, i.e., }(A+B) and f(a+b) arc of the same affection. The same result follows from (3) also. 3.17. Delambre's analogies. We have sin (A + B)=sin 1A cosB +cos A sin 1 B. Substituting for sin A, cos B, etc., their equivalents from Art. 3.8, we get sin 3(A+B)=sin (8—b)+sin (8-a) sin s sin (s-c) sin a sin b sin c sin (s-b) + sin (8-a) cos JC sin c 2 sin fc cos (a - b), cos fc. sin c * Napier (1550-1617) discovered these analogies and published: them in his Mirifici Logarithmorum Canonis Descriptio in 1614.

·60 SPHERICAL TRIGONOMETRY sin 3(A+B) cos C cos }(a - b) cos jo LUULUU ... (1) Similarly, sin 1(A-B) cos 10 sin 3(a - b) sin dc . (2) ... cos (A+B) sin 10 cos f(a+b) cos c ... (3) and cos (A-B) sin C sin f(a+b) sin c (4) The above four formulae are known as Delambre's analogies and were obtained by him in 1807, though published afterwards in Connaissance des Teme., 1809, p. 443. Sometimes they are improperly called Gauss's Theorems,* 3,18. Napier's analogies can easily be obtained from those of Delambre. Thus dividing (1) by (3) and (2) by (4) we get the first two analogies of Napier. Similarly dividing (4) by (3) and (2) by (1) we get the other two analogies of Napier. Delambre's analogies * According to Professor Simon Newcomb (1835-1909) these analogies were first published anonymously by Delambre (1749- 1822) although Gauss (1777-1855) was the first to use them in Spherical Astronomy. Gauss published them in Theoria motus corporum coclestium in 1809 and Mollweide in Zach's Monatliche Correspondenz in 1808.

·ANALOGIES OF NAPIER AND DELAMBRE 61 also may be obtained from those of Napier. Thus squaring the first analogy of Napier, we have cos? (a - b) tan? (A + B)= cot? 10. Cus? (a + b) Adding 1 to both sides, we get sec2 1(A+B) cos? }(a - b)cos2 C + cos2 }(a + b)sin? 10 cos' f(a + b) sin10 }{1+cos (a - b)}cos? |( + {1+cos (a + b)}sin2 } cos 1(a + b)sina 10 1(1+cos a cos b+sin i sin b cos (") cog" }(a + b)sin? 10 *(1+cos c) cos2 /(a + b)sin? 10 cos2 °C cos2 {(a + b)sin? :( whence cos 1(A + B) cos 1(a+b) sin fc cos C which is the third analogy of Delambre. Other apalogies can also be obtained similarly. 3.19. Deduction of the analogies of Napier and Delambre. We have from Art. 3.4 sin A sin B sin b sin c sin c sin a sin c so that sin A +sin B sin a + sin b (1) 2 ... sin c ·62 SPHERICAL TRIGONOMETRY Again we have from Ex. 2, p. 39 sin (A + B) sin C cos a + cos b 1 + cos C (2) And from the polar triangle of ABC, we get (Ex. 1, p. 53) (sin a + b) cOS A +cos B 1-cos C (3) sin c Hence sin a + sin b sin c sin A + sin B cOS A + cos B sin C 1- cos 0 sinc sin (a + b) or tan (A + B)= cos (a-b) cos (a + b) cot 10, which is Napier's first analogy. Again sin a + sin b cos a + cos b sin A +sin B sin C sin c 1+ cos C sin C sin (A + B) or tan (a + b)= cos 3(A-B) cos 7 (A+B) tan c, which is the third analogy of Napier. On taking the negative sign in (1) the other two analogies are obtained in a similar manner. Next consider the colunar triangle A"BC where All is the point diametrically opposite to A. For this triangle A and a are unaltered and the other ·ANALOGIES OF NAPIER AND DELAMBRE 63 parts are changed into their supplements and (2) becomes (Ex. 9, p. 43). sin (A-B (4) 1- cos COS a ... sin , - COS C and from the polar triangle of A"BC, we get (Ex. 2, p. 55) sin (a - b) sin c cos B-cos A 1+cos C (5) Multiplying (1) by (5) we get sin A +sin B sin C cos B-cos A 1+cos C sin a +sin b sin(a−b) sin c sin c or, sina f(A + B) sin (A - B) cosa fC sin C sin a + sin b sin? C sin (a-b), or, sin? 3(A+B) cos2 FC cos2 /(a−b) cos2 10 by (4), which is the first analogy of Delambre. Similarly multiplying (1) by (3) and dividing by (2) we get cos (A - B)-_ sin f(a+b) sin C sin C which is the fourth analogy of Delambre. On taking the negative sign in (1) and multiplying it by (3) and (5) respectively, we get the remaining two analogies of Delambre.

·64 SPHERICAL TRIGONOMETRY EXAMPLES WORKED OUT Ex. 1. If a spherical triangle is equal and similar to its polar triangle, shew that (1) Dec a=bec b sec c+tan b tan c, (2) sec? A + gec2 B + sec? C + 2 sec A sec B sec C=1, (Science and Art Exam. Papers.) (1) We have cos a=cos 6 cos c + sin b sin c cos A, by Art. 3.1 =cos 6 cos c+sin b sin c cos (-a) =cos 6 cos ( - sin b sin c cos a for A=A'= T-a. Dividing both sides by cos a cos b cos c and transposing we get sec a=sec b sec c + tan b tan c. (2) We have cos a=cos b cos c+sin b sin c cos A, or cos (-1)=cos (1-B) cos (7-C) + sin (1-B) sin (7 -- C') cos A, for a=7-A'=9 - A, etc. Hence -COS A=cos B cos (+ sin B sin Ccos A, or, - sec B sec (=scc 4 +tan B tan , or, sec? A + sec2 B sec? C + 2 sec A sec B gec ('=tan? B tan2 C sec2 B sec? C-gec2 B-sec2 C +1, whence sec? A + sec2 B + sec? C + 2 sec A sec B gec C=1. Ex. 2. Three great circles are drawn through a point P on the surface of a sphere, cutting the sides of the spherical triangle ABC and making with them the angles 01, 91, 41; 02, 02, 42 and 03, 03, 43 respectively. Shew that cos , COS 01 COS V1 cos 02 cos 02 CO8 42 0. сов өз сов ф3 cos 43 ·EXAMPLES 65 Let the three arcs out the side a at the points L, M and N, and let PL and PN make the angles a and B with PM. Then from the triangle PLM, we liave by Att. 3.11 A Yi 42 021 क P 43 09 2 0 1 N B M L с cos PM cos e + cos a cos PML sin a sin PML cos 0, -cos a cos Og sin a sin 8.2 Again from the triangle PMN, we have cos PM= - cos 03 + cos B cos Oy sin B sin 0.2 Hence (cos 0, -cos a cos 03) sin B=sin al- cos 03 + cos B cos 02) Or, Similarly, and cos o, sin B + cos 03 sin a=cos 02 sin (a + b). cos di Bio B+cos 03 sin a=cos 02 sin (a +B), Cos y sin B + cos 43 sin a=cos 2 ein (a + B). Hence eliminating sin a, sin B and sin (a + b) from the three equations, we get cos, COS 01 COS 1 cos 02 cos 02 cos 42 = 0). cos 03 cos 03 cos 43 5 ·66 SPHERICAL TRIGONOMETRY Ex. 3. If two great circular arcs are drawn from the vertex C of a spherical triangle ABC, one perpendicular on AB and the other bisecting the angle C, and o be the angle between them, shew that cos ila-b) tan (A-B). cos f(a + bi (Dublin Univ. Exam. Papers.) tan o Let the perpendicular and the bisector meet AB at D and E respectively. From the triangle CAD, we have by Art. 3.11 cos A COS CD sin (10-0)' and from the triangle CBD, we get cos B cos CD= sin (1C +0) Thus COS A cos B sio (10-0) siu (C+0) Hence cos A-cos B COS A + cos B sin (IC-)-sin (C++) sin (10-0) + sin (7C + )' or, tan 1(A + B) tan (A - B)=tan o cot C. Hence substituting the value of tau 3(1 + B) from Napier's first analogy (Art. 3.16) we have cos (a-6) tan (A-B). cos /(a+b) tan o Note.-Substituting the value of tan }(4-B) from Napier's second analogy we get sin da-b) tan 1(A + B). sin (a + b) tan o= ·EXAMPLES 67 Exc. 4. If 8 be the longth of the arc through the vertex of an isosceles triangle, dividing the base into segments a and B, shew that tan ja tan 8-tan f(a + 8) tan ja -8), where a is one of the equal sides of the triangle. ¡C.U., M.A. & M.Sc., 1934.) Let ABC be an isosceles triangle having AC=BC, and let 8 meet the base AB at D. Then from the triangle ADC, we have by Napier's third analogy (Art. 3.16) tau 3(a +8) cos (D-A) tanta, cos }(D+A) where D represents the angle ADC. Again from the triangle BDC, we have by Napier's fourth analogy tan ](a-8)=8 sin }(1-D-B) sin }(1-D+B) tan IB cos I(D + A) cos }(D-A) tan 1B. Hence in ultiplying, we get tap 1(a + 8) tan Ila-5) = tan fa tan 1B. Ex. 5. If a, b, c, d be the sides of a spherical quadrilateral taken in order, and ' the diagonals, and othe arc joining the middle points of the diagonals, shew that cos o cos a + cos b + cos c + cos d 4 cos 78 cos 18 ·68 SPHERICAL TRIGONOMETRY Let the diagonals meet at P and let E and F be their middle points. Let PC and PD be denoted by x and a so that PA=:-* and PB=8'-'. Let the angle APB be 0. Then cos arcos (8-2) cos (8'-') + sin (8 - x) sin (8' - x') cos e, =cos (8'-') cos x-sin (8' - ') sin x cos 6, cos b = cos c= cos x cos x' + sin x sin x' cos e, and cos d == cos (8 - x) cos x'- sin (8 - x) sin x' cos 0. Hence cos a + cos + cos c+cos d -{cos (8 - x) + cos x}{cos (8'- *') + cos x'} + cos o{sin(8 - x) - sin x}{sin(8'—x') -sin x'} =4 cos 18 cos d' cos (18 - x) cos (18' - «') +4 cos 8 cos 18 cos 48' sin (18-x) sin (18'-'). Again from the triangle PEF, we have oos = cos (18) cos (18'—x')+sin (18 - 1) sio (18'- x') cos 8. Therefore cos a + cos b + cos c + cos d=4008 © cos 18 cos 18', or, COS cos a + cos b + cos ( + cos 1. 4 cos focos 18 EXAMPLES 1. In any spherical triangle, shew thai cos a tan B+cos b tan A + tan ('=cos a cos b tan A tan B tan C. 2. In any spherical triangle, shew that sin b sin c + cos b cos c cos A=sin B sin C-cos B cos C cos a. (Cagnoli.) (Dacco Uni., 1932.)

·EXAMPLES 69 3. Prove that 2 cos f(a+b) cos }(a,h) tan 1c=sia b cos A +sin a cos B, and tan }(A + a) tan (B-u)=tan }(A-a) tan ](B+b). 4. If A=2, shew that tan ja= tan Ab tan c 1-tan fb tan c (Science and Art Exam. Papers, 1899; Dacca Uni., 1930.) 5. Shew that in an equilateral triangle og rin 1:1 + log cos fa + log 2=0. 6. If A and A' denote the angles of an equilateral triangle and its polar reciprocal, shew that Cos Acos A' =cos A +cos A'. (Science and Art Exam. Papers.) 7. In any triangle, shew that cos a sin l-gin a cos b cos C COS 1 gius 2 sin (a + b) sin? 10 and cos 4 + cos B sin c 8. Prove that cos (B-1) cos (4-0) tan }u-tan }l cos (-tan tc cos 13 tan tb-tan sa cos (-tan tc cos A 9. Shew that tan c= cos A cot a + cos B cot b cot a cot b-cos A cos B ·70 SPHERICAL TRIGONOMETRY 10. In a spherical triangle, show that 1+cos a-cos b-cos C sin (S-A) 4 cos ta sin fb sin c and 1-cos A +cos B + cos C cos (s-a)= 4 sin 1A cos fB cos JC 11. Shew that cos fc=cos jla-b) cos1C + cos2 }(a + b) sind fC, and sin> {c=sinja-b) cos? 10 + sin? }(a + b) sin? 10. 12. Shew that tan? fc tan2 ja-2 tan ja tan fb cos C +tan? 16 1 + 2 tan fa tan fb cos C+tan? ja tan? 16' 1-tan? ja [Substitute for cos a and 1 + tan2 ja in the formula of Art. 3.1.] 2 tan ta for sin a, etc., 1+tan? 10 13. Shew that sin }(B-C) Stan ja -0. sin B sin JC [Substitute values for tan fa, etc.) 14. If 8 and 8' denote the lengths of the internal and external bisectors of the angle C of a spherical triangle and terminated by the side AB, shew that cot 8 = cot a + cot b 2 cos C and cot si cot a cot b 2 sin 10 16. If 81, 82 and 83 denote the bisectors of the internal angle of a spherical triangle, shew that cot o cos JA + cot 02 cos JB+cot 83 cos /C=cot a + cot b + cot c. (Dacca Uni., 1930.)

·EXAMPLES 71 16. If 81, 82 and 8: denote the bisectors of the external angles of a spherical triangle, sbew obat cot 8'1 sin JA + cot 8'2 sin }B + cot ', sin IC=0. 1 17. If s and s' are the seg nents of be base made by the perpendicular from the vertex C, and o end of those made by the bisector of the vertical angle, shew that tan tan2 am tan 2 2 (Dublin Univ. Exam. Papers.) 18. If a ship be proceeding uniformly along a great circle and li, lo and lz be the latitudes observed at equal intervals of time, in each of which the distance traversed is s, shew that SI COS sin }('1 + lz) cos }(1-13) sin 12 denoting the radius of the Earth. 19. If o denotes the angle between the bisector of the vertical angle C of a spherical triangle and the perpendicular from C on the base AB, shew that tan sin (a-0) sin (a + b) cot C. 20. If in any spherical triangle C=A+B, shew that 1-cos acos b + cos C= 21. If in any spherical triangle a+b=1 +c, shew that 1+cos A +cos B -COS C=0. 22. If in a spherical triangle b+c=, shew that sin 2 B + sin 2C=0. (Dacca Uni., 1932.)

·72 SPHERICAL TRIGONOMETRY 23. If A, B, C and D are four points on the surface of a sphere and is the augle between the arcs AB and CD, shew that cos AC cos BD-C08 AD cos BC=sin AB sin CD cos 0. (Gauss.) 24. If a, b, c and d be the sides of a spherical quadri. lateral taken in order, 8 and 8' be the diagonals, and 41 and 42 be the arcs joining the middle points of the opposite sides a and c, b and d, shew that cos b + cos ( + cos 8 + cos 8 CO8 41 4 con fa cos C cos u + COS ( + cos 8 + cos 8' and cos 12 4 cos fl) cos id 25. If one side of a spherical triangle be divided into four equal parts, and 01, 02, 03 and 04 be the angles subtended at the opposite corner by the parts taken in order, then sin (0, +0.2) sin 02 sin 04 =sin (03 +04) sin 0, sin 03. 26. In an isosceles triangle ABC, each of the base angles are double the vertical angle: shew that cos a cosa=cos (c+la), where a is one of the equal sides of the triangle. (London Unirersity E:ram. Papers.) 27. If a, b, c and d be the sides of & spherical quadrilateral taken in order, and 8 and 8' be the diagonals intersecting at an angle 8, shew that cos a cos c-cost cog d CO3 = siu sino 28. Il 41 and 42 be the arcs joining the middle points of pairs of opposite sides a and c, b and d of a spherical quadrilateral, and o the arc joining the middle points of the diagonals 8 and 8, shew that cos 41 cos fa cos dc + cos 42 cos fb cos td-cos o cos 18 cos f8' = }(cos 8 + cos 8').

·CHAPTER IV RIGHT-ANGLED T.:IANGLES 4.1. Formulae connecting the parts of a rightangled triangle. Let ABC be a spherical triangle right-angled at , and let () be the centre of the sphere. At A draw the tangents 4D and 11 to the arcs AB and AC respectively. They lie in the planes AOB and AOC. Let them meet OB and OC produced at D and E respectively. Join ED. A E с B D Since the angle (is & right angle, the planes 004 and OCB are perpendicular to each other.

·74 SPHERICAL TRIGONOMETRY Also the radius OA is perpendicular to both the tangents AD and AE, and therefore the angles OAD and OAE are right angles and OA is perpendicular to the plane ADE. Also any plane through OA is perpendicular to the plane ADE. Hence the plane OCA is perpendicular to the plane ADE. Thus both the planes ADE and OCB are perpendicular to the plano OCA, and so DE, their line of intersection, is perpendicular to the plano OCA. Therefore the angles OED and AED are right angles. Now ОА OD OA OE OE OD that is, cos c=cos a cos b. ... (1) sin a Again sin A: DE AD DE OD OD AD sin c' that is, sin a=sin i sin c. ... (2) Similarly, sin b=sin B sin c. ... (3) tan b Also cOS A= AE AD AE 0A OAAD tan c or, tan b=cos A tan c. ... (4) Similarly, tan a=cos B tan c. (5) ...

·FORMULAE FOR RIGHT-ANGLED TRIANGLES 75 tapa And tan A= DE AE DE OE OE“ АЁ sin b' that is, tan a=tan A sin b. (6) ... Similarly, tan b=tan R sin a. (7) ... Multiplying together (6) and (7) we get tan a tan b tan A tan B sin a sin b cos a cos b 1 1 COS C or, cos (=cot A cot B. (8) Again dividing (2) by (5) we get sin A sin COS a COS C cos a cos b, cos B COS B so that cos B=sin A cos b). (9) Similarly, from (3) and (4) we have cos A=sin B cos a. (10) The above ten formulae * will enable us to obtain the value of any element of a spherical triangle when two other elements (other than the right angle) are given. All the above formulac could be deduced from those of the previous chapter by putting C=. These formulae were known to the Hindu Mathematicians and were used by them to solve spherical right-angled triangles. See A. Arneth - Geschichte der reinen Mathemat !. Stuttgart, 1852. Nasir ed-din al-Tusi (1201-1274) of Persia collected these formulae into a consistent whole in 1250.

·76 SPHERICAL TRIGONOMETRY 4.2. Some important properties. Since cos c = cos a cos b, it follows that either only one cosine is positive or all of them are positive. Hence in a right-angled triangle either two sides are greater than quadrants and one side less than a quadrant or all the three sides are less than quadrants. tan a Again since tan A= it follows that tan A sin b and tan a are of the same sign. Henco A and a are either both greater than or both less than , i.l's, A and a are of the sime affection. Similarly B and b are of the same affection. ) 4.3. Napier's Rules of Circular parts. Napier has given two rules which include in them all the ten formulae established in Art. 4.1. He takes the two sides which include the right angle, the complement of the hypotenuse and the complements of the remaining angles, and calls these the circular parts of the triangle. Thus if C be a right angle, the five circular parts are a, b, 41-(, -A and T-B. He takes a circle and divides it into five sectors and writes one circular part in each sector in the order in which they naturally occur in the triangle. * These rules are due to Napier, and were published by him io his Mirifici Logarithmorum Canonis Descriptio in 1614. He calls them theorems, and while he verifies them in the ordinary way, by testing each of the known relations between the parts of a right-angled spherical triangle, he exbibits their true character in relation to the star pentagon with five right angles.

·NAPIER'S RULES OF CIRCULAR PARTS
77
Selecting any one of the five parts, and calling it
the middle part, the two parts contiguous to it are
fr-B
fr-c
-A
called the adjacent parts and the remaining two are
called the opposite parts. Thus if i-c is taken as
the middle part, then 1-4 and -B will be
adjacent parts and a and b the opposite parts.
Napier's Rules are the following: -
(1) sine of the middle part=product of the tan.
gents of the adjacent parts.
(2) sine of the middle part=product of the
cosines of the opposite parts.
For example,
sin (7-c)=tan (1-A) tan (17-B),
i.e.,
cos (=cot A cot B,
which is formula (8) of Art. 4.1.
Again sin (4* -c)=cos a cos b
or,
cos c=cos a cos 6,
which is formula (1) of Art. 4.1.

·78 SPHERICAL TRIGONOMETRY For a proof of the above rules see Napier's Mirifici Logarithmorum Canonis Descriptio, 1614, pp. 32-35. 4.4. Quadrantal triangle. When one side of a triangle is a quadrant, it is termed a Quadranlal triangle. Evidently it is the polar reciprocal of a right-angled triangle, for if C= 17, we have c'=7-0 = 37. Hence the formulae for a quadrantal triangle are obtained from those of a right-angled triangle by changing the sides and angles into the supplements of the angles and sides. Thus we have the following formulae when the side is quadrant:- с 8 ... ... ... cos ( +cos 1 cos B=0. sin A=sin a sin C. sin B=sin b sin C. tan A +cos o tan (=0. tan B + cos a tan C=0. tan A=tan a sin B. tan B=tan b sin A. cos ( + cot a cot b=0. cos b = sin a cos B. cos a=sin b cos A. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) ... . 4.5. Trirectangular triangle. When all the threc sides of a spherical triangle are quadrants, it is called a Trirectangular or Triquadrantal Triangle. Evidently, all its angles are also right angles (Ex. 5, p. 26). Thus in a trirectangular triungle the sides and the ·DIRECTION ANGLES AND DIRECTION COSINES 79 angles are all right angles. Each vertex is the pole of the opposite side, and consequently the arc joining u vertex to any point in the opposite side is a quadrant. Since the angle between two radii of the sphere is equal to the aru joining their extremities, it follows that the radii fron the centre of the sphere to the vertices of a trirectangular triangle are mutually at right angles. Thus in the figure of Art. 4.7 the radii OA, OB and OC are mutually at right angles. Direction Angles and Direction Cosines of 4.6. a point. The angles which the radius to a point on the surface of the sphere makes with the radii to the vertices of a trirectangular triangle, at the centre of the sphere, are called the Direction Angles of that point, and the cosine of these angles, the Direction Cosincs of that point. Thus taking ABC lo be a trirectangular triangle and P any point on the surface of the sphere whose centre is O, we have the angles POA, POB and POC as the direction angles and cos POA, cos POB and cos POC as the direction cosines of the point P. Since the arcs P.1, PB and PC measure the angles which OP inakes with 04, OB and OC, we can define direction angles as the angular distances of a point on the surface of a sphere from the vertices of a trirectangular triangle on it. Thus the arcs PA, PB and PC are the direction angles and ·80 SPHERICAL TRIGONOMETRY cos PA, cos PB and cos PC are the direction cosines of the point P. Since the angles POA, POB and POC remain the same for all positions of P on the straight line OP, their cosines also remain the same. Thus we get the idea of Direction Cosines of the line OP referred to three rectangular axes 04, OB and OC in solid Geotietry. 4.7. Theorem. If any point P on the surface of a sphere be joined to the vertices of a trirectangular triangle ABC by grcal circular arcs, then will cos2 PA + cos2 PB + cos2 PC=1. i.c., the sum of the squares of the direclion cosines of a point on the surface of the sphere is cqual to unity. P с B D We have by Art. 3.1 cos PA=cos AB cos PB + sin AB sin PB cos ABP. =sin PB cos ABP, since AB is a quadrant.

·THEOREM 81 Similarly, cos PC=sin PB cos PBC=sin PB sin ABP. Hence, squaring these and adding, we have cos?PA+ cos2PC=sin2PB=1-cos? PB. Thus cos2PA+ cos2PB + cos2 PC=1. P:3 Cor. If Pi, P2 and ba the perpendiculars from the point P on the sides of the triangle ABC, then will sinºPi + sinºP2+ sin?p3=1. 4.8. Theorem. If any two points P and Q on the surface of a sphere be joired to the vertices of a trirectangular triangle ABC by grcat circular arcs, then will cos PQ=cos P A cos Qil + cos PB cos QB +cos PC cos QC. A E F M Nt @ PC L B D From the triangle PAQ, we have by Art. 3.1 cos PQ=cos PA cos QA+sin PA sin QA cos PAQ. 6 ·82 SPHERICAL TRIGONOMETRY Now cos PAQ=COB(PAC-QAC) =cos PAC cos QAC+sin PAC sin QAC =cos PAC COB QAC + cos PAB cos QAB: Therefore cos PQ=cos PA cos QA +sin PA sin QA(cos PAB cos QAB +cos PAC cos QAC). But cos PB=sin PA cos PAB, cos QB=sin QA cos QAB, cos PC=sin PA cos PAC, cos QC=sin QA cos QAC: Hence cos PQ=cos PA cos QA+cos PB cos QB +cos PC cos QC. This theorem expresses the distance of any two points on the surface of the sphere in terms of their distances from the angular points of a trirectangular triangle. . Cor. If P1, P2, P3; 91, 92, 93 be the perpendiculars from the points P and Q on the sides of the triangle ABC, then will cos PQ=sin P, sin qı + sin P, sin 9, +sin P3 sin s.

·DIRECTION COSINES OF THE POLE 83 4.9. If we put l, m, n and V, mi, ml for the direction cosines of P and Q, and 6 for the angular measure of the arc PQ, the two preceding articles give two well-known results of Solid Geometry, viz., (1) 12 + m2? + n2=1 and (2) cos 0=11' + mi' + nn', the direction cosines of OP and OQ being with reference to the three rectangular axes OA, OB and Oc. 4.10. Direction Cosines of the Pole of the Aro joining two points on the Surface of the Sphere. Let P and Q be two points on the surface of the sphere and let H be the pole of the arc PQ. Then by, Art. 4.8, we have cos HP =cos HA coB PA +cos HB cos PB +cos HC cos PC, and cos HQ=cos HA cos QA +cos HB cos QB +cos HC cos QC. But the arcs HP and HQ are quadrants, hence cos HA cos PA +cos HB cos PB + cos HC cos PC=0 and cos HA cos QA +cos HB cos QB+cos HC cos QC=0.

·84 SPHERICAL TRIGONOMETRY Solving the above equations, we get cos HA cos PB cos QC -cos PC cos QB cos HB cos PC cos QA - cos PA cos QC COS HC cos PA cos QB -cos PB cos QA = { }' cos2HA + cos2HB + cos2HC 1 (cos PB cos QC -cos PC cos QB) 2 sin PQ Thus cos HA sin PQ=cos PB cos QC -cos PC cos QB, cos HB sin PQ=cos PC cos QA -cos PA cos QC, cos HC sin PQS =cos PA cos QB -cos PB cos QA. EXAMPLES WORKED OUT Ex. 1. In a right-angled triangle, iſ 8 be the length of the arc drawn from C perpendicular on the bypotenuse AB meeting it at D, shew that (1) sin’d=tan AD tan BD. (2) tan?a=tan BD tan c and tap?b=tan AD tad c. (1) We have from the triangle ACD tan AD=tan ACD sin 8, by (6) of Art. 4.1. Similarly, tan BD=tan BCD sin 8. " This is obtained from the identical relation (mn' -m'n)2 + (nl' -n'])2 + (!m' - I'm)2 = (12 + m2 + n2)(1/2 + m'? + n/2) - (ll' + mm' + nn')?.

·EXAMPLES 85 Hence multiplying, tan An tap BD=sin28 tan ACD tan BCD =gin28: 1.€., sine of the perpendicular is the geometric mean between the tangents of the segments of the hypotcnuse. (2) We have from the triangle BCD, by (5) of Art. 4.1 tan a tan BD tan c tana Hence tan? a=tan BD tan c. Similarly, tan: b=tan AD tan c, i.e., tangent of a side is the geometric mean between tangents of the adjacent segment and the hypotenuse. cos B Ex. 2. Perpendiculars are drawn from the vertices A, B, C of any triangle, meeting the opposite sides at D, E, F respectively : shew that tan BD tan CE tan Af=tap DC tan EA tan FB. (Dacca Uni., 1932.) A FA E с BY D Let the perpendiculars meet at the point P. We have from the triangles BPD and CPD, by (6) of Art. 4.1 tan BD=tan BPD sin PD and tan DC-tan CPD sin PD.

·86 SPHERICAL TRIGONOMETRY Therefore tan BD tan DC tap BPD tan CPD Similarly, tan CE tan EA tan CPE tən APE and tan AF tan FB tan APF tan BPF" Hence multiplying both sides and noting that BÊD=AÑE, CÔD=AF and CỘE-BÊF, tan BD tan CE tan AF we get 1. tan DC tan EA tan FB Ex. 3. OX and 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 aro perpendicular to AB from 0, making the angle COX(=a) with ox. PM and PN are arcs perpendicular to OX abd OY respectively : shew that if OM=x and ON=y, cos a tan x + sic a tan g=tan p. Y A C N Y X M Let OP be denoted by 8 and the angle POC by 0.

·EXAMPLES 87 From the right-angled triangle POM, we bave by (4) of Art. 4.1 tan a cos POX= tan 8 Similarly from the triangle PON, we have tan 11 cos OY =sin POX. Hence tan 8 ten x cos a + tan y sin a=tan : (cos a co3 POX + sin a sin POX) =tan 8 cos 8. But from the triangle POC we have tan p C080 tand Hence cos a tan x + sin a tan y=tan p. EXAMPLE 1. If ABC be a triangle in which the angle C is a right angle, prove the following relations- (1) 2n=sin a sio b. (2) 2N = sin a sin B=sin A sin b. n (3) =gin c. N (4) sina sin?b=sina+sin-sinc. (5) 2 sin' fc=sin? Ja + b) + sin fla-b). (6) sin a tan JA-sin b tan |B=sip (a-b). sin (s-a) (7) tan jB= sin s sin (c-b) (8) tan' JA sin (c + b) (9) tan f(A + B) tan f(A-B)= sin(a−b) ein (@+b) (Dacca Uni., 1930.)

·88 SPHERICAL TRIGONOMETRY 2. In a triangle iſ C be & right angle and D the middle point of AB, show that 4 cosa fc sin?CD=sin?a + sinºb. 3. If o be the length of the arc drawn from C perpendicular to the hypotenuse AB, shew that (1) cos'8=Cos? A + cos2B. (2) cot'= cot?a + cotab. (3) sinº8 sin?c=sin( + sin'b-sin'c. 4. If 8 be the length of the arc drawn from C perpendicular to AB in any triangle, show that c). cos 8 =cosec c (cos'a + cos2 - 2 cos a cos b cos (Cal. Uni. M.A. and M.Sc., 1926.) 5. If the side c of a triangle be a quadrant and 8 be the length of the arc drawn at right angles to it from C, shew that, (1) co328 =cos?a + cos2b. (2) cot? 8 = cotA +cot?B. (3) sin?r=cot & cot 0, where 8 and o are the segments of the angle C. 6. If the side c of a triangle be a quadrant, shew that (1) cos (S-A) cos (S-B)+cos (S-C) cos S=0. (2) tan a tan b + sec C=0. (3) 2 cos (S-A) cos (S-B)=sin A sin B. 7. In the triangle ABC if (= 90°, show that cos a + cos h sin (A + B) 1 + cos a cos b (Cal. Uni. M.A. and M.Sc., 1931.) and sin (A-B)= COB b- -COB a 1-cos a cos b (Dacca Uni., 1931.) 8. If C= 90°, shew that tan Se-cot ja cot Ib.

·EXAMPLES 89 9. If one of the sides of a right-angled triangle be equal to the opposite angle, shew that the remaining parts are each equal to 90'. 10. If 8 be the length of the bisector of the hypotenuse AB of the right-angled triangle ABC, sbew that sin 28 sin'a + aip?b 4 cos 11. Show that the ratio of the cosides of the segments of the base, made by the perpendicular from the vertex, is equal to the ratio of the cosines of the sides. 12. Shew that the ratio of the cosines of the base angles is equal to the ratio of the sides of the segments of the vertical angle made by the perpendicular drawn from it to the base. 13. If aj, ag; B1, B2 and 71, 72 be the segments of the sides of a spherical triangle made by the perpendiculars from the opposite vertices, shew that cos a cos 81 con Yi=cos ap cos Ba cos Y - 14. If P1, P2 and 13, P4 depote the perpendiculars from the base angles A and B to the internal and external bisectors of the vertical angle C, shew that sin Pı sin P3 + sin pa sin P4-sin a sin b. 15. If 1, u and y denote the perpendiculars from the vertices of any triquadrantal triangle on a trunsversal to the sides, shew that sin?+ ain's +sin'v=1. 16. ABC is a spherical triangle each of whose sides is & quadrant, and P is any point within the triangle : sbew that cos PA cos PB cos PC + cot BPC cot CPA cot APB=0 and tan ABP tan BCP tan CAP=1.

·90 SPHERICAL TRIGONOMETRY Solution of right-angled triangles. 4.11. We have seen that a triangle has six parts, three sides and three angles, and the formule established before shew that if three parts are given, we can determine the remaining three parts, and thus completely solve the triangle. In solving numerical examples, we shall have to make use of logarithmic tables. Six cases present themselves. In these cases the right angle forms & known part and we require to know only two other parts. The angle C is taken to be a right angle in all the following cases. 4.12. Case I. Having given two sides a and b. The remaining elements A, B and c are obtained from the formula (6), (7) and (1) of Art. 4.1 cot Arcot a sin b, cot B cot b sin a, COB C=cos a cos b. The solution is unique and the triangle is always possible. EXAMPLE Given a=56°18', b=39°27'; solve the triangle To find c, we have cos c=cos a cos b, or, 10 + L cos C= L cosa + L cos b, or, L cos cs 9.6430438. 63° 55' 21".

·SOLUTION OF RIGHT-ANGLED TRIANGLES 91 To find A, we have . 10+ L cot A=L ont 55°18' + L sin 39°27', or, L vol.1-9.643428C. A = 66° 15' 6". To find B, we bave 10+ L cot B=L cot: 39°27' + L sin 55°18', or L cot B=9.9996157. B=46° 1' 31". 4.13. Case II. Haring given two angles A and B. The remaining elements a, b and c are obtained from the formulæ (10), (9) and (8) of Art. 4.1 cos A= = cos a sin B, cos B=cos b sin A, cos c=cot A cot B. Here also a, b and c are uniquely determined. EXAMPLE Given A = 64°15' and B=48°24'; solve the triangle. We have L cos a= =9.7641507 a=54° 28' 53", L cos b=9.8676405 b=42° 30' 47", and L cos c=9•6316912 C=64° 38' 38".

·92
SPHERICAL TRIGONOMETRY
4.14. Case III. Having given the hypotenuse c
and one side a.
We have from (2), (5) and (1) of Art 4.1
sin A
sin a
sin c
COS B
tan a
tan c
COSC
cos b =
COS a
and .
The elements B and bare determined without
ambiguity, but sin A admits of two values between 0
But since a and A are of the same affection,
i.e., they are either both acute or both obtuse, we take
that value of A which is of the same affection with a.
Thus A is also uniquely determined. The triangle is
thus possible.
If a and c are both quadrants, then A is a right
angle, but b and B are indeterminate.
4.15. Case IV. Having given the hypolenuse c
and an anglc A.
We have from (2), (4) and (8) of Art. 4.1
sin a=sin A sin c,
tan b=cos A tan c,
cot B=tan A cos c.

·SOLUTION OF RIGHT-ANGLED TRIANGLES 93 Thus B and b are uniquely determined, and as a and A are of the same affection, a is also uniquely determined. Thus the triangle is possible. If A and c ara toth right angles, then a is a right angle, but b and B are indeterminate. 4.16. Case Y. Having given one side b and the adjacent anglu A. The formula for determining a, B and care (4), (6) and (9) of Art. 4.1 tan b tanca > cos A tana=tan A sin b, cosB =cos b sin A. Thus a, B and c are uniquely determined. 417. Case VI. Having given one side a and the opposite angle A. Here we have from (2), (6) and (10) of Art. 4.1 sin (1 sin c= 9 sin A sin b=tan a cot A, COS 4 sin B= COS (1 Here c, b and B are to be determined from their sines, and between 0 and 7 there are in general two angles having a given sine. Thus we get two values for each sine, and we expect six different triangles ·94 SPHERICAL TRIGONOMETRY with the given data. But this is not the case. We must have a and A of the same affection, and since sin c must be less than unity for c lies between 0 and 7, sin a must be less than sin A, and so a must be less than A when they are both acute or greater than A when they are both obtuse. Otherwise the solution will be impossible. When this condition is satisfied, we get two values for c, and since cos c=cos a cos b, we get one value for 6 for each value of c, and one value for B, because b and B are of the same affection, which is otherwise evident from the relation cos c=cot A cot B. Thus we see that there will be in general two triangles with the given parts. We say in general because if a and A are equal but not right angles, we have b, B and c all right angles and thus we get only one triangle. In this case A is the pole of BC. When a and A are right angles the solution becomes indeterminate. That we should have two triangles is apparent from the fact that the triangle ABC and its colunar triangle A"BC satisfy the given data, for A=A" and BC is common. If A=a, we get one triangle, for the triangle A"BC is symmetrically equal to the triangle ABC.

·APPLICATION OF NAPIER'S ANALOGIES 95 EXAMPLE Given a=51°20', A=62°12' and C=90°; solve the trianglo. To find c, we have L sin c=10+ L sin a L bin A =30+ 9.8925-9.9467=9.9458. Hence C= 61° 58' cr 1182'. To find b, we have L sin b=L tan a + L cot A ---10 10.0968 +9•7220-10=9:8188. Hence b=41° 13' or 138° 47'. To find B, we have Lasin B=10+ L cos A-L cos a =10+ 9.6687-9.7957 =9.8730. Hence B=48°17' or 131°43'. 4.18. Application of Napier's analogies in the solution of right-angled triangles. Napier's analogies can profitably be used in solving right-angled triangles in the three following cases. Firstly, when the sides a and b are given; Secondly, when the angles A and B are given; Thirdly, when a and B, or b and A are given. and EXAMPLE Solve the triangle having given a=64° 30', b=48° 12' and C=90°. To find c, we have cos c=cos a cos b, or, L cos c= L cos 64° 30' + L cos 48° 12'-10 =9.6340 +9.8238-10=9.4578. Hence c=73° 19' 28".

·96 SPHERICAL TRIGONOMETRY To find A and B, we have from Napier's firat analogy tan }(A + B)= cos fla-b) cos }(a+b)' or, L tan f(A + B)=10+ L cos 8° 9'-L cos 56° 21' 10+ 9.9956 -9.7436 =10 2520, Honce Similarly (A + B)= 60° 45' 40". L tan }(A - B)=10 + L sin 8' 9'-L sin 56° 21' =10 +9:1516-9.9204 = 9:2319. Hence (4-B) = 9° 39' 53". A= 70° 25' 33" and B=51° 5' 47". 4.19. Solution of oblique-angled triangles. As in the case of right-angled triangles, six different cases present here also, and when we are given any three of the parts, we can determine the remain. ing three parts by making use of some of the formule of Chapter III. We shall not go in details but finish this chapter by giving an application of Napier's analogies to solve an oblique-angled triangle. EXAMPLE Solve the triangle having giren A=130° 5' 22:41", B=32° 26'6'41" and c=51° 6' 11.6". From Napier's third analogy, we have L tan fla+b)= L cos }(4-B) - L cos (A+B) + L tan }c =9*81844-9:18158 +967950 =10 31636. Hence (a + b)=64° 14' 7".

·SOLUTION OF OBLIQUE-ANGLED TRIANGLES 97 Similarly L tan 3(a−b)= 1, sin }(A - B)-L sin }(A + B) + L tan jo =9:87663-9:99433 + 9•67950, whence 10-h)=200'22". a=84 14' 29" and b=44° 13' 45'. To find C, we use Delambre's third analogy, whence L sin }(= L cos }(A + B)- L cos }(a + h) + L cos de =9:18158-9.63816 + 9.95529. Hence C=18°22'43". or, ('=36°45'26". The value of (can also be obtained from Napier's first analogy. TXAMPLES. Solve the following triangles having giren 1. a=37° 48' 12", Ans. A=41° 55' 45", a=54° 16', Ans. A=68° 29' 53". 3. A = 36', Ans. a= = 20° 51' 18:5", 4. a= 59° 28' 27", Ans. b=48° 39' 16", or b=131' 20'44", 5. A=23° 27', Ans. (=600, 6. a=138° 4', Ans. A = 142° 11' 38", 7. A=46° 45', Ans. a 44° 53' 904, b=59° 44' 16", B = 70° 19 '15", h=33° 12', B=38° 52' 26", B= 60°, b=31° 43' 3", A = 66° 7' 20", B=52° 50'20", B=127° 9' 40", B=7° 15' b=15° 56', b=109° 41', B=120° 15' 57", C=75° 40', b=69° 32' 55", C=90°. c=66° 32' 6". C=90°. (=60° 44' 40". (=90°. (=37° 21' 38.5". C=90'. = 70° 23' 42". c=109° 36' 18". c=74° 29'. C=153° 44'. c=90°. C=113° 28' 2". C= 90°. B=75° 15' 22"

·CHAPTER V
PROPERTIES OF SPHERICAL TRIANGLES
5.1. Relations between the aros joining three
doints on a great circle and any other point.
Theorem. If Y, Y, Z bc three points on a great
circle, and P any other point, on the sphere, then will
cos PX sin YZ+cos PY sin ZX + cos PZ sin XY=0... (1)
and
cot PX sin YPZ + cot PY sin ZPX
+ cot PZ sin XPY=0...(2)
Р
z
X
Y
X1
Y,
From the triangles PXY and PYZ we have
cos PX=cos PY cos XY +sin PY sin XY cos PYX
and cos PZ=cos PY cos Y Z + sin PY sin YZ cos PYZ.

·RELATION BETWEEN THE ARCS
99
But cos PYX=
cos PYZ ; hence
cos PZ=cos PY cos YZ-siu PY sin YZ cos PYX.
Eliminating cos TYY from these two equations,
we bave
(cos PX -cos PY cos XY) sin Y Z
+(cos PZ-cos PY cos YZ) sin XY=0,
or, cos PX sin YZ-cos PY sin XZ+cos PZ sin XY=0.
Writing XZ=-ZX, by measuring arcs in one
direction as positive and, in the opposite direction, as
negative, we have
cos PX sin YZ+cos PY sin ZX + cos PZ sin XY=0 ... (1)
Again by Art. 3.10 we have from the triangles
PXY and PYZ
sin PY cot PX=cos PY cos XPY +sin XPY cot PYX
and
sin PY cot PZ=cos PY cos YPZ + sin YPZ cot PYZ.
Multiplying these two equations by sin YPZ and
sin XPY respectively and adding, wo have
sin PY (cot PX sin YPZ +cot P2 sin XPY)
=cOS PY sin XPZ,
or, putting XP2=-2PX, we have
cot PX sin YPZ + cot PY sin ZPX
+ cot PZ sin XPY=0... (2)

·100 SPHERICAL TRIGONOMETRY 5.2. Particular cases. (1) Median. If Y be the middle point of X2, then PY is the median of the triangle PXZ, and (1) gives cos PY: (co3 PX + cos PZ) sin XY sin XZ > or, cos PY cos PX + cos PZ cos XY + cos Y Z Thus if m be tbe length of the median bisecting the side a of the triangle ABC, we have * COS m= cos b + cos C 2 cossa (2) Internal Bisector of an angle. If PY bisects the angle P, we have from (2) cot PY=sin XPY (cot PX+cot PZ) sin XPZ cot PX + cot PZ cos XPY + cos YPZ Thus the internal bisector 8 of the angle A of the triangle ABC is given by 1 cot = (cot b + cot c). 2 cos JA Gudermann, Niedere Sphärik, $ 400.

·SPHERICAL PERPENDICULARS 101 (3) External Biseotor of an angle. If PZ bisects externally the angle XPY, then Y+z=JT-3 XPY 8nd XPz=3m + 3 x Y, so that cot PY-cot PX cot PZ= 2 sin } XPY Thus the exterpal bisector d' of the angle A of the triongle ABC is given by 1 cot s'= (cot b-cot c). 2 sin jÀ (4) If XZ be a quadrant, we have COS PY= =cos PX sin YZ+cos PZ sin XY. Thus if the base BC be a quadrant, and a point D be taken in it, we have = cos c sin DC + cos b sin BD. COS AD= 3.3. Spherical Perpendiculars. Let the Arcs PX, PY and PZ when produced meet another great circle at right angles at the points X, Y, and 2, respectively, then P is the pole of the great circle X,Y,21, and each of the arcs PX , PY, and PZ, is a quadrant. (See fig. of Art 5.1.) Hence cos PX=sin XX1 cos PY=sin YY1 and cos P2=sin 221, nd (1) of Art. 5.1 becomes Bin XX, sin YZ + sin YY, sin ZX+ sin 22 sin XY=0 (3) ·102 SPHERICAL TRIGONOMETRY Similarly (2) of Art. 5.1 yives tan XX, sin YPZ+tan YY sin ZPX +tan 22, sin XPY=0 (4) ... Since the angle between any two arcs PX and PY is measured by the intercept made by them on the great circle X,Y121, i.e., by X,Y, (Art. 1.8), we get tan XX, sin Y,2, +tan YY, sin 2,X1 +tan 22, sin X, Y,=0 ... These are the relations connecting the spherical perpendiculars XX1, YY, and Z2 from the points X, Y and Z on the great circle X,Y,Z). 5.4. Theorem. If three arcs meet at a point, the ratio of the sines of the arcs drawn from any point on one of the arcs, perpendicular to the other two, is constant. Let OA, OB and OC be the three arcs and let a and B be the lengths of the perpendiculars from a point P in OB on the arcs OA and OC respectively. Then from the two right-angled triangles, we have sin a sin OP= sin AOP sin e sin COP or, sin a sin ß sin AOP which is constant, sin COPY for it is independent of the position of P on the arc OB.

·CONCURRENCY OF THREE ARCS 103 Conversely, if froin any point Plin OB, perpendiculars a' and ßl are drawn on 01 and OC so as to satisfy the relation sip a sin al sit B si ß then P' will lie on the great circle through and P, namely OB. 5.5. Concurrenoy of three arcs. Theorem. If three arcs joining a given point with thi angular points of a triangle meet the opposite sidos, the product of the sines of the alternate segments of the sides arc equal. А E с B' Let the arcs joining A, B and C with the given point P meet the opposite sides in D, E and F. respectively.

·104 SPHERICAL TRIGONOMETRY Then from the triangles APF and BPF we have sin FB sin AF sin AP sin APF sin AFP and sin BPF sin BFP sin BP 80 that sin AF_sin AP sin APF sin FB sin BP sin BPF Similarly sin BD sin DC sin BP sin BPD sin CP'sin CPD' and sin CE sin EA sin CP sin CPE sin AP sin APE Hence multiplying the corresponding sides of the three equalities and noting that BÊD=AÑE, CÔD=AF and CỘE=BÊNF, we have sin Af ein BD sin CEO sin FB sin DC sin EA 1. The corresponding theorem for a plane triangle is Ceva's theorem. * 5.6. The converse theorem can also be easily proved. Several theorems on concurrency of arcs are immediately deducible from it. Thus The perpendiculars drawn from the vertices of a spherical triangle to the opposite sides meet at a point. + * See Russel's Pure Geometry, Chap. I. + Gudermann, Niedere Sphärik, $68; Schulz, Sphärik II, $47.

·CONCURRENCY OF THREE ARCS 105 The bisectors of the angles of a spherical triangle meet at a point.* The arcs joining the angular points of a spherical triangle with the middis points of the opposite sides meet at a point. 6.7. Theorem. If three arcs passing through the vertices of a triangle be concurrent, the products oj the sines of the alternate segments of the angles of the triangle atequal. Let the arcs AD, BF and CF meet at P and divide the angles A, B, C of the triangle ABC into the segments A1, A2 ; Bu, B, and C1, C2. (See fig. of Art. 5.5.) Then from the triangles ABD and ACD, we have sin c sin BD sin 41 and sin DC sin b sin A2 sin ADC sin ADB Hence sin BD sin 4 sinc sin DC sin Ay 6 sin sin a Simil«rly sin CE _sin B1 sin EA sin B ginc 2 and sin AF sin PB sin C, sin b sin C, 'sin ai * First proved by Menelays. 7-A

·DOOOOOO.COM
106
SPHERICAL TRIGONOMETRY
1
sin A, sin B sin C
"herefore
1.
sin A, sin B, sin C,
This is also another criterion for the concurrency
ni three arcs.
The converse case can also be easily proved.
5.8.
Conoyolic points.
Spherical transversal.
Theorem. If a great circle intersects the sides of
a triangle ABC at the points L, M and N, then will
sin AN sin BL sin CM
sin NB sin LC sin MA
1.
M
z
X
B
M
N
Draw AX, BY and CZ perpendiculars on the
great circle LMN. Then from the triangle ANX we
hava
sin AX=sin AN sin ANX.

·CONCYCLIC POINTS 107 Similarly from the triangle BNY, we have sin BY=sin NB sin BNY. sin AN Hence sin AX sin NB sin PY' Similarly sin BL sin CL sin BY sin CM sin CZ and sin CZ sin MA sin AX Hence multiplying and writing - sin LC for sin CL, we have sin AN gin BL sin CM sin NB sin LC sin MA -1. This theorem along with its analogue for plane triangle was obtained by Menelaus.* Its converse is also true, namely if three points L, M, N be taken on the sides of a triangle satisfying the above relation, then they will lie on a great circle. Note 1.-Any transversal must cut either one or all the three sides of the triangle externally. Thus the arc LMN cuts only the side BC externally whereas the arc LM Ni cuts all the sides externally. Hence there will always be the negative sign. Note 2.--Several formulae for right-angled triangles are easily deducible from Menelaus' theorem. Thus if C=90° and AN and AM are quadrants, then L will be the pole of AC and the theorem becomes cos c = cos a cos b. Again the triangle NBL with AC as transversal gives sin a=sin A sin c. Other formulae are similar. ly obtained by taking any three arcs as forming a triangle with the forth one as the transversal. * In Greek geometry this theorem is known by the pame of Regula Sex Quantitatum. See Sphaerica by Menelaus or Des Claudius Ptolemaus Handbuch der Astronomie by Karl Manitius, Bd. I, pp. 45-51.

·108 SPHERICAL TRIGONOMETRY 5.9. Theorem. The great circle bisecting the sides of a triangle intersects the base in points which are equidistant from the middle point of the base. X 2 pe D Let ABC be the triangle and let D, E and F be the middle points of the sides BC, CA and AB respectively. Draw the secondaries AX, BY and cZ on EF. Let EF and BC when produced meet at the points P and Pl. Clearly these are two diametrically opposite points. Now in the triaugles AFX and BFY, we have AF=FB, AÊF=BÝF and Afx=BFY. Hence the triangles are equal in all respects so that AX=BY. Similarly AX=CZ. Therefore AX=BY=OZ. Again from the equality of the triangles BPY and CPZ, we easily get BP=CP',and as BD=CD, DP=DP'ra quadrant. we have

·CASEY'S THEOREM
109
5.10. Casey's Theorem.* If two points P and Q
be taken on the surface of a sphere, of which is
within a spherical triangle ABC, and if 2nı, 2ng and
2nz be the sincs of the triangles QBC, QCA and QAB,
then
nicos PA + ng cos ?B +ng cos PC=n cos PQ.
P
C.
B
Join P and Q to the points A, B, C.
Produce AQ to meet BC in F. Join PF and PQ.
Then since B, F and C lie on a great circle, and P
is any other point, we have by Art. 5.1
cos PB sin FC + cos PC sin BF=cos PF sin BC.
Similarly for the points A, Q, F and P, we have
cos PA sin QF+cos PF sin AQ=cos P Q sin AF.
Dr. Casey, Spherical Trigonometry, p. 81.

·110 SPHERICAL TRIGONOMETRY from these two Hence eliminating cos PF equations, we get cos PA sin QF sin BC + cos PB sin FC sin AQ +cos PC sin BF sin AQ =cos PQ sin AF sin BC. If QD be drawn at right angles to BC, then sin QD sin QF = sin F so that sin QF sin BC=sin QD sin BC_2n, by Ex. 4, p. 40. sin F sin F' Similarly sin AQ sin FC= 2ng sin 4Q sin BF sin F sin F 2123 3 > 2n and sin AF sin BC sin F Hence we have 12cos PA +ng cos PB +13 cos PC=n cos PQ. 5.11. Normal co-ordinates of a point. If from a point P perpendiculars 2, B, y are drawn to the sides of & triangle ABC, then sin «, sin B and sin y are called the Normal co-ordinates of P with respect to the triangle Normal co-ordinates are clearly analogous to trilinear co-ordinates with respect to a plane triangle.

·NORMAL CO-ORDINATES 111 If 2n, 2n, and 2ng be the sines of the triangles PBC, PCA and PAB, we have sin a sin a=2nį, sin ß sin h ==212 and sin γ sin c=2ng. When the rating of the co-ordinates are known, the point is determined. EXAMPLE Find the normal co-ordinates of the point where the perpendi. culars from the angular points to the opposite sides meet. A F E с BS D Let the perpendiculars AD, BE and CF meet at P. Now from the triangles ABD and ACD, we have by (9) of Art. 4.1 cos B=cos AD sin BAD, and cos C=cos AD sin CAD. cos B Hence sin y sin BAD sin CAD by Art. 5.4. cos C sin B Similarly, cos C COS A sin a sin y Hence i.e., sin a cos A=gin B cos B=sin y cos C sin a, sin B and sin y are respectively proportional to cos B cos C, cos C cos A and cos A cos B.

·112 SPHERICAL TRIGONOMETRY 5.12. Normal co-ordinates with respect to a trirectangular triangle. Their fundamental properties. We have seen (Art. 4.5) that the arc joining the vertex of a trirectangular triangle to any point in the opposite side is a quadrant. Hence if P be any point on the sphere, and D, E, F the points where AP, BP and CP meet the opposite sides, then PD, PE and PF will be complementary to AP, BP and CP respectively. (See figure of Art. 4.7.) Now the Normal co-ordinates of P are sin PD, sin PE and sin PF. Hence with respect to the trirectangular triangle they are cos AP, cos BP and cos CP, and these are generally represented by l, m and n. In fact l, m, n are the direction cosines of OP referred to threc rectangular axes OA, OB and OC, O being the centre of the sphere. They satisfy the following properties-- (i) 12 + m2 + n2=1 and (ii) I! + mm' + nn' = cos PQ, 1, m, n and I', m', n' being the normal co-ordinates of two points P and Q on the sphere. EXAMPLES 1. IfD be any point in the side BC of the triangle -130, shew that cot AD sin BAC =cot AC sin BAD + cot AB sin DAC. 2. If two sides of a spherical triangle be supplementary, prove that the median passing through their intersection is a quadrant. (R.U.I., 1895.)

·EXAMPLES 113 3. The medians of a triangle ABC intersect at P and meet the opposite sides at D, E, F respectively : show that (0)' sin PA : sin PD :: 2 cos já : 1, : sid PP ; sin PE :: 2 oon jb :1. sin PC : din PP :: 2 cos jc : 1. 4. From any three points on a great circle, secondaries 2, 4, 2; x', y', z' and a", y", 2" are drawn to the sides of a triangle : show that gid 8. sin y, sin z sin ', sin y', sio z' O. sin 2", sin y", sio z" 6. Three points P, Q and R lie on a great circle, and X, Y and 2 are three otber points on the sphere : sbow that cos PX, cos PY, cos PZ cos QX, cos QY, cos QZ cos RX, cos RY, COB RZ 6. If the bisectors of the angles of the triangle ABC meet at P, sbew that sin BPC sin AP Bid CPA sip BP sin APB sin CP sip a : sin b : sin c. (1) sin? AP: sin? BP: sin2 CP sin (8-a) sin a sin (8-0). sin b sin (s-c) sin c 8 ·114 SPHERICAL TRIGONOMETRY, 7. Find the Normal co-ordinates of the point where the aros joining the angular points of a triangle to the middle points of the opposite sides raeet. Ans. Proportional to sin B sin C, sin C sin A and sin A sin B.' 8. If the internal bisectors of the angles of the triangle ABC intersect at P and meet the opposite sides in D, E and Pres pectively, shew that sin PD sin PE ein PF sin a sio AD ein b sin BE sin o sin CF 1 {sin? s +sin s sin (8-a) sin (s-6) sin (s-c)}} (R.U.I., 1896.) 9. If a, a'; B, B' and y, y' be the segments of the perpendiculars to the sides of a spherical triangle drawn from the opposite vertices, sbew that tan a tan a'=tan B tan B'tan y tan w' and cos (a ta'; COB & COB & COB (B + B) COB B cos b' cos (y + gy's cos y cos y 10. ABC is a spherical triangle, E is the middle point of BC, and AD is drawn at right angles to BC ; shew that tan ED sin (B+C)-tan ja sin (B-C). (Sci. and Art., 1894.)

·CHAPTER VI SPHERICAL AREA OF SPHERICAL TRIANGLE. EXCE3S 6.1. Area of a spherical triangle. Girard's theorem.* А B A" Let ABC be a spherical triangle. Produce the. sides AB and AC. They will meet at A" where A" is the point diametrically opposite to A (Art. 2.11). Thus we get a lune ABA"CA with the angle A. Similarly BC and BA produced give the lune This theorem is due to Girard and was published by him in 1629 in his Invention nouvelle en Algèbre. A rigorous proof of it was given by Cavalieri in his Directorium generale uraniometricum in 1632.

·116 SPHERICAL TRIGONOMETRY BCBAB of the angle B, and CA and CB produced give the lune CACHBC of the angle C. The triangle ABC forms a part of each of these three lunes. Let r be the radius of the sphere. Then ABC + ABC=lune ABANCA=2Ar2 ABC + ABC=lune BCBAB=2Br2, ABC + ABC"=lude CAC"BC=2Cr2, by Art. 2.11. aun Now the triangle ABC" is antipodal to ABC and hence they are equal in area (Art. 2.12). Hence putting A"B"C in place of ABC" and adding the three equalities above, we get 2 triangle ABC + area of hemisphere=2 (A+B+C)-2, or, triangle ABC + r2 = (A + B + C)?. Therefore area of the triangle ABC=(A + B + C--)r2 ... (1) The expression A+B+C-7 is called the Spherical Excess of the triangle ABC and is denoted by the symbol E. It measures the excess of the sum of the angles of a spherical triangle over the sum of the angles of a plane triangle (both being expressed in circular measure) and hence the name. If we put 2S=A+B+C, we get SEE + 1.

·AREA OF A POLYGON
117
Cor. 1. If Ej, E, and E, be the spherical
excesses of the colunar triangles of ABC on the sides
a, 6 and c respectively, then
E,=2A-E, E,=2B-E, and E,=2C-E,
and their areas are
(2A-E), (2B-E) 2 and (2C-E).
Cor. 2. The sun of the areas of any triangle
and its colupar triangles is equal to half the area of
the sphere.
6.2. Area of a Polygon. Take a polygon of n sides
and let denote the sum of its angles. Take any
point within the polygon and join it to all the angular
points. Then the polygon is divided into n triangles
and its area is equal to the sum of the areas of the n
triangles. Hence
area of the polygon=(sum of the angles
of the n triangles-na);
=(+21-17)p2={I-an-2)}2
= Era,
where E is the spherical excess of the polygon.
Cor. Area of a spherical quadrilateral is
(A + B + C + D-2) 2
6.3. Girard's theorem enables us to get the area
of the spherical triangle when the sum of the angles
are linowa. When the three sides or two sides and the

·118 SPHERICAL TRIGONOMETRY included angle are given, the relations established in the following articles will enable us to find the area. 6.4. Cagnoll's theorem.* To shew that sin E- {Bin 8 sin(8-a) sin (8-6) sin (8-c)} 2 cos da cos fb cos fc We have sin E=sin (S-2)=-cos S sin {(A + B) sin dc-cos }(A + B) cos {C. Hence substituting the values of sin }(A + B) and cos }(A + B) from Delambre's analogies (Art. 3.17), we get sin E= sin £C cos } {cos }(a−b) - cos f(a+b)} cos dc sin C cos c sin ja sin }b 2n sin a sin b sin ja sin b cos c by Art. 3.3 n (2) . .. 2 cos la cos to cos fc 6.8. Expressions for cos LE and tan E. To shew that 1+cos a +cos b + cos c COB JE= 4 cos da cos i cos 2c and tan E= 2n 1 + cos a + cos b + cos C * Cagnoli, Trigonometria, $ 1146. See also Lexell, Acta Petro." politana, 1782, p. 68. For a geometrical proof see Art. 6.11 below.

·EXPRESSIONS FOR COS E AND TAN JE 119
We have
cos JE=cos (8-37)=sin 8
= sin f(A+B) cos + cos }(A+B) sin fc
= {cos IC cos }(a−b)+sina 16 cos }(a+b)} sec fc
by Delambre's analogies Art. 3.17
= {cos da cos fb + sin ja sin } cos C} sec fc
(3)*
cos? La cos2 16+sin ja cos fa sin }b cos fb cos C
cos ja cos fb cos fc
(1 + cos a)(1 + cos b)+sin a sin b cos C
4 cos la cos to cos fc
1+ cos a + cos b + cos C
4 cos fa cos tb cos gc
...
...
Hence dividing (2) by (4), we have
tan LE
2n
1+cos a + cos b + cos C
(5)1
* Lagrange, Journal de l'E'cole Polytechnique, Cahier, 6;
Logendre, Géométrie, Note 10. Gudermann, Niedere Sphärik,
$ 162:
+ Euler, Acta Petropolitana, 1778. For a geometrical proof
soo Art. 6.11 below.
Do Gua, Mémoires de l'Académie des Sciences, Paris, 1783,

·120 SPHERICAL TRIGONOMETRY 6.6. Formulae for Colunar triangles. Let E, be the spherical excess of the colunur triangle A"BC. If an, bz, Cl be the sides and Ali B, C, the angles of this triangle, we have B=2A-E and aq=a, b;=r-b. Ci=-C. A=A, B,=r-B, C,=r-C. Also 8=1-(8-a), 81-Q,=1-8,8,-6,=8-C and 81-C1=8-b, so that n1=n. Now sin sen ni 2 cos ja, cos fb, cos c1' whence by substituting the values of an, 61001 we have n sin JE, = sin (A - 4E) ...(6) 2 cos ja sin b sin c Similarly.. sin fEg= sin (B-4E)= ...(7) 2 sin fa cos jb sin fc and n n sin JE, = sin (C-E) (8) 2 sin ja sin fb cos C 1+cos Q , + oos b, + cosa, 4 cos fa,cos b, cos cu Again cos 11 ·L'HUILIER'S THEOREM 121 whence COB b cosc 1+ocsa cos E, cos (A - E) ..(9) 4 cos ża sin fb sin fc with similar expressions for cose, and cos JE,. Also tan JE,=tan (A- ;E)= 2n cos b-Cos C . (10) 1 + cos a with similar expressions for tan JE, and tan JE,. It should be noted here that E1, E, and E3 being spherical excesses are necessarily positive, and each of them is less than 21. (Art. 2.9.) Hence A-E, B-EC-E are each less than T. 6.7. L'Huilier's theorem, * To shew that 7) tan ļE={tanje tan (8-a) tan}(8-b) tan}(8-c)}. We have tan te sin (A + B + C cos (A+B+C - ) sin }(A + B) –sin 3(1-C) cos (A+B) + cos }(1-C) sin }(A+B)-cos C cos (A+B) + sin C * See Legendre Géométria, Note 10. See also Grunert's Archiv der Math. und Physik., XX, 1863, p. 358 for Gont's proof of L'Huilior's theorem.

·122
SPHERICAL TRIGONOMETRY
sin C
cos }(a-b) - cos de cos JC
cos }(a + b) + cos C
by Delambre's analogies,
sin }(8-b) sin } (8-a) sin : sin(8-0)
Cos 18 cos }(8-0) sin (8 a) sin (8 - b)
by Art. 3.8.
=v{tan 18 tan }(8-a) tan }(8-6) tan }(8-c)}...(11)
6.8.
The Lhuilierian.
We have by (11)
tan 11
v{tan 18, tan}(8,-a,)tan }(81-bl)
tan j(81-C1)},
whence
tan 4(2A --E)=^{cot 18 cot }(8-a) tan :(8-5)
tan :(8-c)}... (12)
Similarly,
tan 1(2B-E)=v{cot j& tan }(-a) cot :(8-5)
tan (8-c;}... (13)
and
tan 4(2C-E)=v{cot zs tan }(s-a) tan ](8 – )
cot }(8-c)} ...(14)
Multiplying together the equations (11), (12), ...(13)
and (14) we get
tan 1E tan (2A-E) tan 1(2B-E) tan 4(2C-E)
=cot 18 tan }(8-a) tan }(8–6) tan ](8-0)=L2 ... (16)

·EXPRESSIONS FOR SIN DE AND COS DE 123 where L is called the Lhuilierian # of the Spberical triangle. Thus > tan E tan (2A-E)= cutts L tan }(8-a)' L tan 38 – b) L tan :(8-0) tan 1(2B-E) = 3 and tan }(2C-E): 6.9. Expressions for sin E and cos E. We have sinº E= }(1-COB LE) { 1 1+cos a + cos b + cos C 4 cos fa cos 16 cos c by Art. 6.5 =={1 -1} cosa fa + cos2 kb + cosa fc-1 2 cos fa cos i cos c 1- cosa fa-cos? }b-cosa 2c +2 cos fa cos 46 cos de 4 cos la cos jb cos dc ... sin få sin 4(8 — a) sin }(8 -- b) sin }(8-). (16) сов }a cos 2b cos с Tho Dame Lhuilierian is suggested by Dr. Casey after tho pame of L'Huilier who obtained this expression.

·124 SPHERICAL TRIGONOMETRY Similarly, =={1 Cos? {E={(1+cos LE) 1+cos a + cos b + cos c 1 to 4 cos da cos }b cos fc cosja + cos2 b + cosa c + 2 cos la cos fb cos de-1 4 cos ja cos th cos c ... costs cos }(8 –a) cos }(8-b) cos }(8 -c) (17) cos sa cos }b cos sc L'Huilier's theorem is obtained by dividing (16) by (17). 6.10. Expressions for sin (2A-E) and oos 1(2A-E). Substituting in (16) and (17) the values of the elements of the colunar triangle A' BC from Art. 6.6, we get, sin? (2A-E) cos to cos }(8—a) sin }(8—b) sin }(8-c) (18) cos ja sin fb sin sc and cos2 (2A-E) sin 48 sin }(s-a) cos }(8 — b) cos }(0-c) (19) cosa sin sb sin c Hence by division, we get tan? (2A-E)=cot 18 cot }(8-a) tan }(8-b)tan 1-c), which is the same thing as (12) of Art. 6.8. ...

·GEOMETRICAL REPRESENTATION
125
6.11. Geometrical representation of the Spherical
Ex0888.
Let D, E and F be the middle points of the
sides BC, CA and AB of the triangle ABC, and let
EF meet BC produced at rand Pl. Then by
Art. 5.9, we have
А
F
X
2
Pl
B
D
PÊY=P.CZ,
PŘY=FÂx and EĈZ=EÂX.
Hence PBY+PĉZ=PẾF+PIĈE-FÂX-EÂX
=2- (A+B+C)=1-E,
80 that PÊY=P:ĉ2=-}E, i.e., complement
of half of the special excess.
Now from the right-angled triangle PBY, we have
sin PBY
sin PY
sin PB
but
PB=47- ja and PY=17-EF ;

·126 SPHERICAL TRIGONOMETRY therefore cos EF cos JE cos fa 1 + cos at cos b + cos C 4 cos da cos fb cos c (Ex. 3, p. 89.) Again cos PBY=sin Pcos PY=sin P sin EF. But from the triangles PBF and EAF, we have sin P sin F sin fc sin PB sin' }b sio EF sin A and sin F so that cos PBY= sin }b sin fc sin A cos ja n Therefore sin fE=cos PBY= 2 cos la cos {b cos fc' which is Cagnoli's formula. EXAMPLES WORKED OUT Ex. 1. In & spherical triangle if cos C- - tan fa tan jb, shew that C= A + B. We have cos C= -tan da tan fb, cos' C= sin ja sin fb cos C cos da cos 36 cos? C 1-cos C sin ja sin fb cos C cos da cos fb +sin ja sin fb cos C tan je cot C, by Arts. 6.4 and 6.5. Hence -cot C-tan Ertan (S-1)--oot S, OI C=S=$(A +B+C). Ca A + B. so that

·EXAMPLES
127
1
Ex. 3. Shew tbat
{sin E sin }(2A – E) sind (XB-E) sin }(20-E)}}
sin :=
2 sin A sin B sin jf.
We have
{sin } E-sin }(2A-E) sin }(2B-E) giu }(20-2)}}
2n2
by (2) of Art. 6.4 and (6), (7), (8) of Art. 6.89
sin a sin b sin c
2 sin & sin (s—a) sin (8-6) sin (8-c), by Art. 3.9
sin a sin b sin c
=2 sin 8 sin JA sin B sin 1C, by Art. 8.8.
Hence the result.
Da. 3. If E' be the spherical excess of the polar triangle, and
E1, E, and Ez those of the colupar triangles, show that
tan ZE'={cot E tan Ei tan E, tan 4E3}.
(Prouhet.)
Let d', b', c' be the sides and A', B', C'the angles of the polar
triangle of ABC, then
E'=A' + B' + O'q=2 (-8),
28'-d'+b+c=2*-E,
s'-'=f('+c'-d')=}(2A-E),
8'-b'=}(d'+a'-63=$(2B-E),
8'--'= (a + b'-c)-1(2C-E).
and
Now tan E'=\{tan fe' tan f(-a) tan fl'-6) tan f(-)}
by Art. 6.7

·128 SPHERICAL TRIGONOMETRY: Henco substituting the values, we have tan 1E'=tan f(-8)=cot je ✓{tan 3(2x-E) tan 7/24 -E) tan 7/2B-E) tan 7(20-E.) ^{cot E tan ZEtan E, tan 1E3). EXAMPLES If E1, E, and Ez be the spherical excesses of the colupar triangles on the sides a, b, and a respectively, shew that 1. sin E sin JE, =tan Jb tan fc. 2. sin E1 tan ta sin E2 tan 16 sin JE: tan Ac sin E tan fa tan lb tan c . 3. sin 1E {sin E sin ? E sin ļE, sin E;} cot sa cot fb cotic 4. tan E cot Ei=tan Is tap }(s-a). tan E cot Eg=tan s tan f(s-b). tan E cot 1E3=tan fs tan f(s-c). 6. cot JE tan E, tan E, tan Ez=cota J$. tan E cot E, tan E, tan 1E3=tan? )(8-a). tan E tan E, cot JEg tap 1E3=tan" }(s-b). tan E tan E, tan E, cot Es=tan his-c). 6. In an equilateral triangle of side a, shew that tan {E =tan fa tan in tan ja. (Dacca Uni., 1930.) 7. In an isosceles triangle shew that tan {E=tan fortan 1(a + c) tan fla- c), where a is one of the equal sides.

·EXAMPLES 129 8. If the angle C of a spherical triangle be a right angle, ahew that (1) sin JE=sin ja sin }b sec jo. (ii) cos JE = cos ja cos b suc fc. sinc gins a sinab COS E + COS C cos a CO3 0 9. If the sum of the angles of a spherical, triangle be four rigb' angles, shew that cog? ja + cos2 1b + cos2 {c=1. 10. A given line is divided into two isosceles triangles, and the area of one of them is n times the area of the other ; shew that n-1 A tan 1A cos 0=tan n+1° 2' where A denotes the angle of the lune and one of the equal sides. (Sci. and Art, 1894 ; C.U., M.A. & M.Sc., 1926.) 11. Shew that sin JE sin JE, sin Exsin Es=N%. 12. Shew that 1E =tan fa tan fb sin C-1 (tan ja tan fb) sin 2C + B and y be the arcs joining the middle points of the sides of a spherical triangle, shew that sin JE=2{sin o sin (o-a) sin (o-B) sin (0-2)} where a + B + =20. 14. If the area of a spherical triangle be one-fourth of the area of the sphere, sbew that the arcs joining the middle points of its sides are quadrants. (London University.) 15. Shew that cot JE=cot C+ cot ja cot }} sin C (C.U., M.A. & M.80., 1927.) 9 13. If a, ·130 SPHERICAL TRIGONOMETRY APPROXIMATE FORMULAE 6.12. Legendre's Theorem.* If the sides of a spherical triangle are small compared with the radius of the sphere, then cach angle of the spherical triangle cxceeds by one third of the spherical excess the corresponding angle of the plane triangle, the sides of which are of the same lengths as the arcs of the spherical triangle. Let a, Band y be the lengths of the arce forming the sides a, b, c of the spherical triengle ABC, so that the circular measures of the sides are a B and r being the radius of the sphere. . T T 19 Then B ข COS COS COS T T cos A = cos a- cos b cos c sin b sinc ß sin sin 7. T T lat 192 21 72 B 183 31 73 Y 18 + 31 p3 mer } { T 1 84 182 21 72 {1- 1 - 1 y2 21 1 41 74 ... 41 T {-1} {* -ost ...) * Legendre, Mémoires de Paris, 1787, p. 998; Trigonometrio, Appendix V. See also Gauss, Disquisitiones generales circa superficies curvas, $ $ 27, 28, and Mertens, Schlomilch's Zeitschrift, 1876.

·LEGENDRE'S THEOREM 131 1 Hence neglecting powers of - beyond the fourth, we have 1 B2 + y2 - 2 1 04-34-71-682,8 24 • C08 A By r2 (1-1983 {$*********-2 min01393" } {1+2***"} + B2+72-02 at +B+ +4 -26272 – 2y2q2 - 2a 239 2By 24Byr (1) If A', B' and Cl be the angles of the plane triangle with the sides a, B and y, we have (Art. 3,6) B2+72-22 cos A = 2By and sin 2A=1-cos? A 2B2y2 + 2y2q2 + 2a288–24 -34-yt 4B272 Hence cos A=cos Al- By sina A1 683 = cos AI- Asin A (2 ) ... 3r3 where A={By sin A', i.e., the area of the plane tri. angle (Art. 3,10).

·132 SPHERICAL TRIGONOMETRY Now if o be the excess of the angle A over the angle A', we have cos A=cos (A' + 0) =cos A-o sin Al approximately, o being a very small quantity. Hence from (2) we have os 4 3r2 Thus A=A+ 3r2 Similarly, B=B' + and C=C! + 3 3r2 8r2 80 that Д A + B + C= A' + B' + C! + = + OL A + B +C-= 4, i.e., E= ... (3) r2 Therefore A= A' + + e, E, B=Bl to 1 3 E and C=C! + 1 3 E.. (4) .... we 6.13. We have seen in Art. 6.1 that the area of the spherical triangle is Er2, and from (3) of the previous article have Erl= A. Thus the areas of the spherical triangle and of the plana triangle with sides of the same length are approxi. mately equal, when the sides are very small as pared with the radius of the sphere. com- ·APPROXIMATE VALUE OF E 133 A closer approximation of the area is given in the following article. value of the spherical 6.14. Approximate excess. * We bave by L'Huilier's theorem (Art. 6.7) tan [E={tan 18 tan 3(8-a) tan (8–b) tan :(8-c)}} Now tan js = Br-ı(8)3 + ... - 1168)2 + fs(1-382 + ...), .) 1-882 + ... . =18(1-482+ ...)(1 - 482 + ...) = 38(1+;582) approximately. Hence tan LE=(1^{1+25*)(–a){1+talo-aje). 360–b}{1+ b(–)}.£6–0){1+36 –ciel]*, =}{818-a)(8-6)(8-c)}} {1+ 8% +(8-a)2 + (8--6)2 + (8 -c)2 12 + } "(8! - )(81-B)(8'--y) 812+(81 – )9 +(8! – B)2 + (8! - y)2 +.... + } 12r2 * Gauss, Disquisitiones, $ 89.

·134 SPHERICAL TRIGONOMETRY where 28'=a + B+y. Thue tan 4E= 1+ 2% +82 +7 2472 . 472 approximately, or, E= ${1+ 92 +82 + y2 2472 } .. ... (5) since the quantities are very small. Hence to this order of approximation, the area of the spherical triangle exceeds that of the plane triangle 1 a* + B2 + y of the latter. If in (5) we neglect by 24 the fourth power of t, we get the result (3) of Art. 6.12. EXAMPLES 1. Shew that a closer approximation for A is given by A = A' + 1 / 2 E+ (B? +2 - 2a"). 180 . 2. Shew that a **(1 :)) sin A B9 789-3a 1+ 1+ sin B 67 60r3 approximately. 3. Sbew that for & closer approximation By ain'A' Byla? - 38? - 37) sin'A' COB A =cos A' 180r* 4. Show that if A- A + 0, then approximately By Bin A a3 +78° +7979 67% 120r + 679 {1+- 2 ·CHAPTER VII CIROLES CONNECTED WITH A GIVEN TRIANGLE INSCRIBED AND CIRCUMSCRIBED CIRCLES. HART'S CIRCLE. 7.1. Inscribed and Circumscribed Circles. Circles can be described touching the sides of a given spherical triangle or passing through its angular points. The contact again may be internal or external, i.e., the circle may be wholly within the triangle or it may be outside the triangle. The circle which can be inscribed within the given spherical triangle so as to touch each of its sides internally, is called its Inscribed Circle or Incir. cle. Its pole will be the point of intersection of the internal bisectors of the angles of the given triangle. Its angular radius will be denoted by the letter r. A circle which touches one side of the triangle and the other two sides produced, is called an Escribed Circle or Excircle. Its pole will be the point of intersection of the bisectors of the external angles. There will be tbree such excircles to a given triangle, and we denote by the letters T 1, rg and is the angular radii of the excircles touching the sides BC, CA and AB respectively. It is evident that the excircle ·136 SPHERICAL TRIGONOMETRY touching BC is nothing but the incircle of the Colunar triangle A"BC. Thus the three excircles are but the incircles of the colunar triangles. The circle which passes through the angular points of the given triangle, is called its Circumscribing Circle or Circum-circle. Its pole will be the point of intersection of the arcs bisecting the sides of the triangle at right angles. Its angular radius will be denoted by the letter R. 7.2. The Incirole, To find the angular radius of the small circle inscribed in a given triangle. F B Let ABC be the given triangle. Bisect the angles B and C by great circular arcs meeting at 1. From I draw ID, IE and IF at right angles to the sides.

·THE INCIRCLE
137
1. Then the triangles IBD and IBF having the
angles at D and F right angles, the angles at B equal
and IB common, are equal in all respects. So also
the triangles ICD and iCE sro equal. Therefore
ID=IE=IF,
and the triangles IAE and IAF are equal, so that AI
bisects the angle A. Thus the internal bisectors of the
angles of the triangle ABC meet at I. A small circle
drawn with I as pole and ID as radius will touch the
sides at D, E and F and will thus be the incircle of
the given triangle.
Now from the triangle IBD, we have by (7) of
Art. 4.1
tan ID=tan B sin BD=tan B sin (s—b),
or denoting ID by T, we have
tan r=tan {B sin (s-b).
Similarly, tan r=tan A sin (8-a)=tan C sin (8-c)
(1)
Again substituting the value of tan }B from
Art. 3.8 we bave
VE
tan r= 1
sin (8-a) sin (8-c)
sin 8 sin (8-5)
sin (8-5) =
n
sin 8
0.0

·138
SPHERICAL TRIGONOMETRY
Similarly substituting the value of the sings in
(1), we get
sin 4B sio 1C sina,
tan 1
cos JA
sin C sin hAsin b,
(3)
cos fB
sin A sin B
sin c.
cos 10
and hence by Arts. 3.13 and 3.14
tant
- cos Scos(S – A) cos (S – B) cos (S-C)}}
2 cos A cos B cos įC.
N
2 cos A cos B cos C.
.. (4) +
...
Again since
cos S +cos (S- A)+cos (S-B)+cos (S-C)
=4 cos A cos fB cos dC
we have
cot ar
{cos S+cos (S-A) +cos (S-B)+cos (S-C)}
2N
(6)
* Loxoll, Acta Petropolitano, 1782.

·THE EXCIRCLE 139 7.8. The Exolrole. To find the angular radii of the escribed circles of the given triangle. B 1, A C Lot ABC be the given triangle. Produce AB and AC to meet at A". Then the circle escribed to the side BC is the incircle of the colunar triangle A"BC, the parts of which are a, -C, A, 5 -B and #-C. If 281 be the sum of the sides of the colunar triangle, we have 8,57- (8 - a), 8, -a=7-8, etc. 7-6, TT 11 Hence if rį be the radius, we have by Art. 7.2, tan =tan A sin (81 –a)=tan A sin 8. (6) Proceeding as in Art. 7.2 or substituting the elements of the colunar triangle A"BC in the formulae of Art. 7.2, we get n tan 11 (7) ... .. sin (8-a)' cos }B cos Csip a, ... (8) ... COBA N 2 cos A sin fB sin c' ... (9) ·140 SPHERICAL TRIGONOMETRY and cotri {-cos S-cos(S - A) + cos(S-B) + cos(5 –C)}. 2N (10) 13 7.4. The rådii 72 and of the other two excircles are easily obtained in the same manner or by appropriate interchange of letters in rj. Thus n tan ra= tan B sin s= sio (8-5)' etc., n and tan T3 = tan JC sin 8= etc. sin (8-c)' 7.5. The Circumcircle. To find the angular radius of the small circle described about a given triangle. E D Let ABC be the given triangle. Bisect the sides BC and CA at right angles at D and E by great ciroular arcs meeting at P. Join PA, PB and PC.

·THE CIRCUÁCIRCLE 141 Then the triangles PBD and PCD, having BDO CD, PD common and the angles at D right angles, are equal in all respects, so ibat PB'=PC. Similarly from the equality of the triangles PCE and PAE, we have PC=PA, so that PA= PB=PC. Hepce a circle with P as pole aud radius PA will pass through the angular points of ABC, and will thus be the circumcircle of the triangle. Now from the triangle BPD, we have by Art. 4.1 tan BD=tan BP cos PBD=tau BP cos (S-A), or denoting the radius by R, we have tan ja=tan R cos (S-A), tan za tan R cos (S - A) Similarly, ... (11) tan c tan R= tan 36 cos (S-B) cos (S-C) Substituting the value of tan la from Art. 3.13 we bave cos S tov tan RE سے cos (S – A) cos (S-B) cos ($ --C) cos S N (12) ...

·142 SPHERICAL TRIGONOMETRY Again since (Ex. 11, p. 56) cos (S - A)= -cos Scot fb.cot fc, we have tan R= tan ja tan }b tan £c. (18) cos S n Also -cos S (Ex, 15, p. 56) 1 2 cos la cos tb cos dc Hence tan Rs 2 sin ka sin }b sin fc (14) * ... n We have from Ex. 14, p. 56. cos (8 - A) sin A cos fb cos ļc сов за so that tap R= sir da sin A cos 16 cos c (15) Again since sin (-a)+sin (8-5)+sin (8-c)-sin s =4 sin ja sin jb sin fc, wo have tan R 1 2n sin (8-a)+sin (8-5)+sin (8-c) -sin s }. (16) .. . Lexell, 1.c. This result follows at once from Ex. 16, p. 56.

·CIRCUMCIRCLES OF COLUNAR TRIANGLES 143 7.6. Circumcircles of the colunar triangles. To find the angular radii of the circumcircles of the three .colunar triangles. Let R, R, and is be the angular radii of the circumcircles of the colunar triangles on the sides a, b and c respectively. The elements of the triangle A'BC are a, rb, 7-C, A, a B and a -C. Hence substituting these values in the formulae of Art. 7.0, we get the formulae for R1, the circumradius of the colunar triangle A"BC, tan fa Thus tan R, = (17) cOS S cos (S-A) N (18) .. tan ja cot tb cot } cos (S-A) (19) 2 sin ja cos 16 cos jo (20) ... sin da sin A sin fb sin fc ... (21) 1 { sin 8-sin(8-a) + sin(8-b) + sin(8-c) D} 2n (22) tan R, tan to Similarly, cos (8-B) N etc., cOS S ·144 SPHERICAL TRIGONOMETRY tań fc and tan Rg= cos (8-C) N etc. 008 S 7.7.: Insoribed and Ciroumscribed oiroles of the Polar triangle. Now I, Let A'B'C' be the polar triangle of ABC. the incentre of ABC, is equidistant from its three sides and hence equidistant from their poles A', B' and C' (Ex. 6, p. 12). Hence IA'=IB=ICI-**-r, i.e., a circle with I as pole and I Al as radius will pass through B' and Cl. Thus, The pole of the incircle of any triangle is also the pole of the circumcircle of the polar triangle, and the radius of the incircle of the triangle is equal to the complement of the circumradius of the polar triangle. Similar reasoning applies to the case of excircles also. Thus the poles of the excircles are the same as the poles of circumcircles of the respective colunar triangles of the polar triangle and the radii of the former are the complements of the respective circumradii of the later. Again since ABC is also the polar triangle of A'B'C', we have the supplemental relation, ·POLAR TRIANGLE 145 The pole of the circumcircle of any triangle is also the pole the incircle of the polar triangle and the circumradius of the triangle is equal to the comple. ment of the radius of the incircle of the polar triangle. It follows from the above that if the radius of the incircle of a triangle is known, the radius of the circumcircle of the polar triangle as also of the given triangle is at once obtained. EXAMPLES WORKED OUT Ex. 1. Shew that (cot rtan R): 1 4nd (sin a + sio b + sin c)* - 1 RE 1 4N3 (sin A + sin B + sin C) - 1. We have from Arts. 7.2 and 7.5 sin 8 cot + tan R= + sin (8-a) + sin ( sb) 2n + sin (s-)-sin s sin 8 + sin (s-a) + sin (8-5)+sin (8-c) 2n 1 sin }(b + c) cos fa +Bin ja cos }(6-c) n Hence equaring both sides, we have 1 (cot + tan R) sin' }(b + c) cosi fa + sinä ja cos}(b-c) + 2 sin ja cob fa sin }(b + c) cos }(b- 16-0} 10 ·146 SPHERICAL TRIGONOMETRY -{ 1. -cos (b + c)} (1 + cos a) + (1 -cos a){1 +008 10-c)} + 2 sin a (sin b + sin c) ** in eo 1 2n8 * 1 + aid a sip b + sin b sin C + sin o sin . - cos a cos 6 cos c 1 (sin a + sin b + sin c)- (1-сos o <cos b-cosC + 2 cos a cos b og c) ܙܘ ܘܗܘ ܀ 1 (sin a + sin b + sin c) -1. sin 6 sin c Again sinco gino sin A (Ex. 7. p. 55) siu B sin C N we have (cot # + tan R)- ano (sin A+ ein B + sin C)-1. 1 Similarly, (cot r, -tan R) = (sin b + sin c-sin a)'-1, 1 (cot r,-tap R)' (sin c + sin a-sin b)'-1, in: 1 and CD (cot, -tan R) (sin a + sin b-sin c)'-1.

·EXAMPLES
147
Ez. 2. Angular distance between the poles of the circumcircle
and the incircle.
.
If d be the length of the great circular arc joining the polos of
the incircle and the circumcircle of a triangle, then will
cos: 8=sin', cos' R + 608% (P.-).
R
P
B
E
D
Let I and P be the poles of the incircle and circumcircle of
the triangle ABC, and let PI be denoted by 8. Through I and P
draw two secondaries to BC meeting it at D and E respectively,
Then we have by Art. 3.7
cos 8=gin ID sin PE + co3 ID cos PE cos ED.
But
BD=8-6, BE=fa; hence ED=}(0-6).
Also
ID=P, sin PE=sin R sin PBE = gia R sin (S-A),
and
COA R
COS PE
cos
Denco
Cos sin r sin R sin (S-A) + cos rcos R
cos fic-0)
OOS

·149 SPHERICAL TRIGONOMETRY = sin 7 sid R sin (S-A)+cos cos Rsin }(B+C) cos JA by Delambre's first analogy (Art. 9.17) gip cos R tan R sin (S-A) + cot , sin (B+C) cos JA sinr cos R cos S sin (S-A) + 2 cns JB con la sin f(B+C) N "} by Arts. 7.2 and 7.5 Binr cos R Bin A + sin B + sin C 2N Therefore we have by Ex. 1, cos- sin, cos? R{(cot r + tan R)2 +1} = sin , cos' R + cos'(R-1). 7.8. Hart's Circle. In the plane geometry we have the well-known theorem of Feuerbach that the inscribed and escribed circles of a plane triangle are all touched by another circle, namely, the Ninepoints Circle. Sir Andrew Hart discovered in 1861 * that the theorem holds in the case of spherical triangles also. He demonstrated that the inscribed circles of & spherical triangle and its colunar triangles are all Seo Quarterly Journal of Mathematics, Vol. IV. p. 260,

·HART'S CIRCLE
149
touched by another small circle. This circle touches
iọternally the incircle of the triangle and externally
the incircles of the colunsr triangles.
7.9. Spherical Radius of Hait's Circle.
Let ABC be the given triangle, and 1, 71, 72, ty
the radii and I, 11, 12, 1; the poles of the inscribed.
and escribed circles. Let p be the radius and H the
centre of Hart's circle. Then since Hart's circle
has internal coutact with the incircle and external
contact with the excircles of ABC, we have
1.35
FA
3
T
H:
1
Hl=p-t, Hl,=p+11, HI,=p+1, and HIZ=p+7g.
Now since the angle A is bisected internally by Al
and externally by Algo they are at right angles to

·150 SPHERICAL TRIGONOMETRY each other. Thus Al is an altitude of the triangle 11.13 Similarly Bl, and CI, are the other altitudes. Let 2v, 2v1, 2v2 and 2vg be the sines of the triangles 1,1,13, II,13, 1131, and II,1,, then 2v=sin I,1, sin Al], 2v2 =sin 131, sin BI, 2v1 =sin 1,13 sin AI, 2v3 =sin 111, sin CI. If IF be drawn perpendicular on AB, we have IF=r, and sin r= sin AI sin JA and sin r , =sin Al, sin 1.4, so that sin r: sin ry=sin AI: sin Alj. 1 Heace . v: Vi 1 sin T sin ti Similarly 1 1 1 v: vi: v2 : V3 1 sin T : sin Ti sin T2 sin T3 Applying Dr. Casey's Theorem (Art. 5.10) on the triangle 1,1,13 we have Vi cos HI, + vg cos H1, +vg cos HI 3 =v cos HI, or, cos (p+1)+cos (p+12) + cos cos (p+13) cos (p-1) Bin T sin 12 sin 73 sia T ·HART'S CIRCLE 151 1.6., cos p (cot 11 + cot r2 +cot 73) - 3 sin p =cos p cott tein p. Thus 4 tan p=cot r , + cot lg + cot T3 - cotr 1 sin (8-a)+sin (s-b)+sin (8-) -sin s n 다 =2 tan R, where R is the circumradius of the triangle ABC. Hence tan p=tan R. 7.10. Angular distance of the pole of Hart's ofrole from the vertices of the given triangle. The lengths of the arcs joining H to A, B and C can be obtained with the help of Art 5.1. Thus applying the theorem to the arc IgAl, we have cos HI, sin Al,+cos Hl, sin Alg=cos AH sin 1913, or, cos (+73) sin Al, +cos (+12) sin Alz =cos AH sin (AI, + AL3). sin Ty But sin Al, sin ? cos ģA sin 413 cOS SA cos Alg=cos T, cos (8.—c), cos Alz=cos T3 cos (8-6).

·152 SPHERICAL TRIGONOMETRY Hence substituting these values in the above equality, we bave sin r2 cos T3 + cos ry sin rg-2 tan p sin ry sin 73 cos AH sin r, cos rg cos ( 86) COS P + sin r3 cos T, cog (8-c) or, cot rg + cot 13 -2 tan p -o) } -o} COS AH {cot cot rg cos (8-5) +cot r, cos (8-c) COS P whence substituting the values of cot ra, cot Tg from Art. 7.4, and of tan p from Art. 7.9 wo get sin (8-6)+sin (8-c) - 2 sin ja sin }b sin fo cos AH sin (8-c) cos (8 – b) COS P + sin (8-b) cos (8-C 0-0}. Hence simplifying, we have COB AH= cos }b cos fc cos p. COS O Similarly cos BH= cos }c cos da cos gb cos p, ·HARI'S CIRCLE 153 and cos CH cos ja cos } cos C CO8 p. 7.11. The lengths iH, Bh and CH can be obtained easily without previous; kuov:ledge of the value of p. Thus from the previous article we have cot ry+cot 13-2 tan p cos 4H { cot r3 cos (8—b) + cotra cos (8-c) COS And applying Art. 5.1 to the arc Alll, we have cos (0-r) sin Al, -cos (p+rı) sin AI =cos AH sin (AI,-AI), which on simplification becomes cotri-cotr-2 tanp COS AH { com cot r cos (8-a)- cot ry cos ..}. COS P cos AH sin a Thus cott2 + cot 73-2 tan p= n cos p co8 AH sina and cot r, -cot 1-2 tan p= n cOS P Hence equating we have tan p= (cot r, +cot ra+cot ng-cot r)=$tan R.

·154
SPHERICAL TRIGONOMETRY
and cos AH=.
п со8 р
sin a
fot
cot ry+cot 73-cot ry+cott
+co7}
cos ab cosc
cos ja
COS p.
Thus the value of p is simultaneously obtained
with tbat of AH.
of the
7.12. Baltzer's Theorem.* The pole
great circle through the middle points of two sides of
a triangle is also the pole of the circumcircle of the
colunar triangle.
А
F
X
E
2
B
с
DG
A'
Draw AX, BY and CZ at right angles to the great
circle EF passing through the middle point E and E
Baltzer, Trigonometrie, $ 5.

·BALTZER'S THEOREM
155
of the sides AC and AB of the triangle ABC. Let
these perpendiculars meet at O'. 'Then O' is the
pole of the great circle EP.
We have by Art. 5.9
AY=BY=CZ=p (say),
then O'A= +p and O'B=0'C=3-p.
Hence O'B=0C=-(O'A- J*)=-O'A=0 A,
where Al is the point diametrically opposite to A.
Thus the point O' is equidistant from the points
B, C and Al, 1.c., the vertices of the colunar triangle
A'BC and hence is the pole of its circumcircle.
7.13. Theorem.it If from a fixed point P on the
surface oj a sphere, a great circular arc be drawn
to cut a given small circle in A and B, then will
tan {PA tan PB=constant.
O'
D
B
• Lexell, Acta Petropolitana, 1782, p. 65.

·156 SPHERICAL TRIGONOMETRY Let o be the pole of the given small circla Druw O'D perpendicular to AB. Then the triangles O'AD and O'BD are symmetrically equal and hence AD=BD. Now from the triangle PO'D, we have cos PO'=cos PD cos O'D, and from the triangle XO'D, we have cos AO'=cos AD cos O'D. Hence cos POI cOS AO cos PD COS AD or, cos AD-cos PD cos AD + cos PD cos A0'- cos POI cos A0+ cos PO' i.e., tan }(PD- AD) tan (PD+AD) =tan }(PO-A0') tan ;(PO' + A0'). Thus tan PA tan PB=tan :(8-p) tan f(8+) =constant, where 8 denotes the angular distance PO and p the angular radius AO'. It is evident that this result does not depend on the positions of A and B, 80 that it holds for all positions of the arc PAB drawn through P. The constant tan (8-p) tan ;(&+p) is called the spherical power of the point P with respect to the circle. It is positive when P is outside the circle, negative when Pis inside.

·LEXELL'S LOCUS 157 7,14. Lexell's loous.* The base and the area of a spherical triangle being given, the locus of the vertex is a small circle. BH B с Let BC be the given base, and B" and Cll be the points diametrically opposite to B and C respectively. Then in the triangle AB"C", the angle BW=-B and C1 = -C. Suppose P to be the pole of the circumcircle of the triangle ABC". Join PA, PB" and PC". Then we have PĒ"C" = PĈ" B", PĈIA=PÂC" and PÂB" =PÊ"A. Therefore B" + CII - A=PBNCI + PC" BII = 2 PÊNCI=2 PCHBA. * Lexell, Acto Peteopolitana, 1781, I, p. 112.

·158 SPHERICAL TRIGONOMETRY Hence if the angle PB"C" or PC"B" is known, the pole P can be determined. Now the area of the triangle ABC is given; hence its spherical excess E is also known. But E = A + B + C-= A +7 - B! +-C1 — == -(B" + C1 - A). 7 Thus B" + CH – A, 1.c., the angle PB"C" or PC"BH is known, so that P is determined and the circumcircle of AB"C" is completely known. As A is a variable point, it follows that the locus of A is a small circle through B" and C1-the circumcircle of the triangle AB"C". EXAMPLES. Prove the following relations for a spherical triangle :- 1. tad tan fı tan ra tan r3=n2. cotr tan rı tan ra tan tz=sin s. tan r cot , tan ra tan rz=sin? (s -a). tan r tan ri cot to tan rz=sin? (s—b). tan , tanrı tan ry cot r3=sin? (s-c). 2. cot R cot Ricot Rg cot Rg=NS. tan R cot R, cot Rg cot Rz=cos S. cot R tan Ry cot Rg cot Rz=cos? (S-A). cot R cot R, tan Rg cot Rz=cos? (S-B). cot R cot Ricot Rg tan R3 = cos (S-C). 8. cot r1: cot ry: cot 13 : cott =sin (s-a): sin (s -b): sin (8-c) : sind, tan R, : tan R, : tan Rg=cos (S-A): cos (8-B) : cos (8-C). ..

·EXAMPLES 159 4. cot 91 + cot tq + cot ry-cot, r=2 tan R. taa ly + tan Rg + tan Rg-tao R=2 cot r. cotr-cot 11 + cot 12 + cot 13 =2 tun Ri. tan R-tan R, + tau lig + van lig = 2 cot f1 ... cotr sin s=cot $fB cot fC. 6. tao R + cot r=-tan R, trot : 1 = tan Rg + cot =tan R3 + cut 73-}cot r + rot 71 + cot rg + cot 73). 7. tan R tan Rj+tan Rg tan Rg=cutr co$ 11 + cot ry cot 73. tan 8. +tan 19+tan 13-tant = }(1 + cos a + cos b + cos c). Cut 91 +cot 99 + cut 73 - COLT 9. tan R + tan R, + tana Ra + tan2 R3 -ī. cut?r + coll 11 + coll ry+Cota 13 10. cos A tan? R + tan Ri-tan? Rg-tan' R3, COL+ COL 1-Cotery-co:? 13 COS O tant cog (S-1) cos (S-B) cos (S-C) 11. tan R 2 cos i cos B cos 0 12. cosecr -cnt (8-b) cot ( sc) + cot (8 -c) cot (s-a) + cot (s-a) cot (8-5). cosecon=cot (s-b) cot (s —c) -cots cot (s-b) - cct scot (s-c), cot (s-a) cot ( sb) cot (s-c) 13. + 2 cot s sin? r sindro sua ng 8:D? 13 es 3 cot (s-a) cot (5-6) cot (8-c). 14. cosec 11 + cosec? 19+ cosec ra-cosec? - 2 cot 8{cot (s-a) + cot (s-b) + cot (8-c)}. 15. V1 + (cut 71-tso R)2 + V1 + (cot re-tan Rj +v1 + (cot 13-lap R): V1 + (cotr + tan R). 16. Shew that in an equilateral spherical triangle ten R=2 tan r.

·160 SPHERICAL TRIGONOMETRY 17. ABC is an equilateral spherical triangle, P the pole of the circle circumscribing it, and Q any point on the sphere : sbew that cos QA + cos QB + cos QC=3 cos PA co3 PQ. (C. U. M. A. & M. Sc., 1926.) 18. If 8 be the angular distance between the roles of the circumcircle and the incircle of a spherical triangle, shew that sin a + sin b tein c ein T sin R 4 sin ja sin fb sin fc' cos o > end Bec R sec, sin' :=tanR-2 tan R tan r. (London Unio. Exam. Papers.) 19. If 8,89 and 8g denote the angular distances betweeo the poles of the circumcircle and excircles of a spherical triangle, shew that cos2 8, =cos) R sin? Ti + cosø (R+;). cos: 0=cos' R sin? Tz + cos? (R + rg). cos? 83 = cos? R sin? 13 + cos” (R + 73). sin8= sin? (R+77)--cosa R sive 11. sin3 89=fins (R+r2! - cosa R 8in) 79. sina 83 = sind (R+13) - cosR sin? 13. 20. If I, 11, 12 and Iz denote the poles of the inscribed and escribed circles of a spberical triangle, shew that cog Ty CO8 73 COB II, : cos II, : cos II; COB Ti cos (s-a) . cos (s-6) Cos ( sc) 21. II S, S1, S2 and $3 denote the sums of the angles of a spherical triangle and its three colunars, sbew that S+S1+Sg+ Sz=3*.

·EXAMPLES 161 32. If P, P1, P, and P, denote the poles of the oircumscribed circles of a spherical triangle and its three colunars, show that Han PPi tan PP, :tan PP: .. = cos ja sin (S--A): cos fb sio (8-B): cos to sin (8-C). 23. If in a spherical «riangle, tie vertical angle be equal to the sum of the base angles, then the pole of the circumcirole will Jie in the base. 24. If ABC be a spherical triangle having each sido a qgam draut, I the pole of the incircle, P any point on the sphere, then will (ccs PA + cos PB + cos PC)2=3 cos& PI. 25. Two circles whose radii are cot-1 a and cot-1-8 touch ex. ternally. Sbow that the angle between their common taogents is 2 cog-1 2 Vaß-1 a+B (C. U. M. A. & M. Sc., 1928.) 26. PAB is a spherical triangle, of which the side AB is fixed, and the les PAB and PBA are supplementary. Prove that the vertex P lies on a fixed great circle. (Science and Art, 1899.) 27. Two circles of angular raŭii, a and B, intersect ortho gonally on a sphere of radius r; find in any manner the area common to the two. (London University.) 28. lf H be the centre of Hart's circle for the spherical triangle ABC, shew that 008 AH : cos BH : cos CH=sect la : sec 16 : 86c% 10. 11 ·: 162 SPHERICAL TRIGONOMETRY 29. If tz, tg and tz be the lengths of the tangents from the verticas A, B and C to Hart's circle, show that cos ti= sec ja cos fb cos fc. cos tq=cos fa sec fb cos fc. cos t3 =cos fa cos fb sec jc. 30. If the side AB of the spberical triangle ABC be intersected by Hart's circle at points distant , and u from A, show that tan 11 = 08 fa-Cos fb cos fc cos tb sin fc and tan tu cos th sin c cos ja + cos fb cos fc circle on the 31. Shew that the intercept made by Hart's side AB is given by 2 tap-1 cos2 fa-cog2 tb 2 cos da cos Jb sin fc }. 32. Shew that the angle between Hart's cirole and a side of the triangle is equal to the difference of the angles of the triangle adjacent to that side. 83. ABCD is a spherical quadrilateral inscribed in a small circle, and the diagonals AC and BD intersect at P: show that tan IPA tan IPC=tan JPB tan IPD.

·EXAMPLES
163
34. ABC is a spherical triangle, and a small circle cuts BC
in P and P', CA in Q and Q', AB in R and R' : shew that
sin Aộ sin AQ”sAB sin AR
cos2 100 cos? JRR
ant
)
sin BP sin BP' ein CQ ein CO sin AR sin AR'
sin CP sin CP sin AQ sip AQ sin BR ain BR
1.
35. P is the pole of the circumcircle of the spherical triangle
ABC, and AP is produced to meet BC in D; shew that if 8 denotes
PD,
sin (R-8)
tan BPD tan JCPD=
sin (R + 8)
If the angle A be a right argle, shew that
cog? R
sin (R-8)
sin (R+8)

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