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A TREATISE ON THE ANALYTICAL GEOMETRY OF THE POINT, LINE, CIRCLE, & CONIO SECTIONS, Containing an Account of its most recent Extensions, With numerous Examples. Price 7s. 6d. A TREATISE ON PLANE TRIGONOMETRY, including THE THEORY OF HYPERBOLIO FUNCTIONS. LONDON: LONGMANS' CO. DUBLIN: HODGES, FIGGIS & CO. A TREATISE ON SPHERICAL TRIGONOMETRY, AND ITS APPLICATION TO GEODESY AND ASTRONOMY, WITH Rummyons xmgws. BY JOHN CASEY, LL.D., F.R.S., Fellow of the Royal Unzversity of Ireland; Member of the Cozunc'l of the Royal Irish Academy; Member of the Mathematical Societzes of London and France; Corresponding Member of the Royal Society of Sciences of Lège; and Professor of th/e Hzgher AIathematics and Mathematzcal Physics zn the Catholic Universzty of Ireland. DUBLIN: IHODGES, FIGGIS, & CO., GRAFTON-ST. LONDON: LONGMANS, GREEN, & CO. 1889. DUBLIN: PRINTED AT THE UNIVERSITY PRESS, BY PONSONBY AND WELDRICK. PREFACE. THE present Manual is intended as a Sequel to the Author's Treatise on Plane Trigonometry, and is written on the same plan. An examination of the Table of Contents, or of the Index, will show the scope of the work. It will be seen that, though moderate in size, it contains a large amount of matter, much of which is original. The sources from which I have obtained information are indicated in the text. The principal are CRELLE'S Journal "< fir die reine und angewandte Mathematik," Berlin, and Nouvelles Annales de Mathématiques, Paris. The examples, which are very numerous (over five hundred) and carefully selected, illustrate every part of the subject. Among them will be found some of the most elegant Theorems in Spherical Geometry and Trigonometry. vi Preface. In the preparation and arrangement of every part of the work I have received invaluable assistance from PROFESSOR NEUBERG, of the University of Liège. For this, as well as for similar assistance previously given in the editing of my Placne Trigonometry, I beg to return that gentleman my most grateful acknowledgments and best thanks. JOHN CASEY. 86, SOUTH CIRCULAR ROAD, DUBLIN. March 25, 1889. CONTENTS. CHAPTER I. SPHERICAL GEOMETRY. PAGE. SECTION I.-Preliminary Propositions and Definitions,. 1 Definitions of Sphere,... 1,, Centre of Sphere,.. 1,, Radius.... 1 Diameter,..... 1 Curve of intersection of sphere and plane,.. 2 Poles of circles defined,... 3 Small circles and great circles defined,.. 3 Intersection of two spheres,....... 3 Secondary circles defined,.... 4 Spherical radius,.... 4 Only one great circle through two points, not diametrically opposite,.... 4 Locus of points of sphere equidistant from two fixed points on sphere,...... 5 Two great circles bisect each other,.. 5 Angle of intersection of two circles on sphere defined,.. 6 ", of two great circles is equal to the inclination of their planes,........ 6 To find the radius of a solid sphere,..... 6 Analogy between the geometry of the sphere and the plane,. 7 Exercises I.,.... 8 SECTION II.-Spherical Triangles,... 9 Spherical triangle defined,....... 9 Correspondence between a solid angle and a spherical triangle,. 9 Lune defined,......... 10 Colunar triangles,.... 10 viii Contents. PAGE. Antipodal triangles,....... 10 Any two sides of a spherical triangle are greater than the third, and the sum of the three sides is less than a great circle,.. 10 Positive and negative poles distinguished,.. 11 Polar triangles defined,...... 11 Relation between polar triangles,.... 12 Sum of angles of a spherical triangle,... 12 Sum of external angles of a spherical polygon,.. 13 Spherical excess defined,..13 Antipodal triangles are equal in area,. 13 Cases in which two spherical triangles on the same sphere have all their corresponding elements respectively equal,.. 14 Properties of isosceles spherical triangles,..15 Area of a lune,........ 15 Girard's theorem.-The area of a spherical triangle,. 16 Ratio of area of a spherical triangle to area of a great circle,. 17 Area of spherical polygon,....... 17 Exercises II.,. 17 Diametrical triangles defined,... 18 Spherical parallelograms defined,..... 18 CHAPTER II. FORMULAE CONNECTING THE SIDES AND ANGLES OF A SPHERICAL TRIANGLE. Three classes of formulae,...19 SECTION I.-First class,. 19 Case I.-Three sides and an angle,.. 19 Exercises III.,... 22 First Staudtian of a spherical triangle,.... 22 Second Staudtian of a spherical triangle,... 22 Norms of the sides and angles of a spherical triangle,. 23 Case II.-Two sides and their opposite angles,. 24 The sines of the sides of a spherical triangle are proportional to the sines of their opposite angles,... 24 Exercises IV.,... 26 Case III.-Two sides and two angles, one of which is contained by the sides,. 27 Exercises V.,...... 28 Case IV.-Three angles and a side,...... 29 Exercises VI.,. 31 Substitutions made in order to pass from a spherical triangle to its polar triangle,. 32 Contents. ix PAGE. SECTION II.-First class continued,... 32 The right-angled triangle,... 32 Comparison of formulie for right-angled triangles, plane an spherical,.... 34 Napier's rules,. 35 Exercises VII.,... 36 Quadrantal triangles,... 38 SECTION III.-Second Class,. 39 Formulae containing five elements,..39 Napier's analogies,... 39 Exercises VIII.,.... 40 SECTION IV.-Third class,...... 40 Delambre's analogies,........ 40 Historical sketch of Delambre's analogies,...41 Reidt's analogies,........ 41 Other applications of Delambre's analogies,. 43 Lhuilier's theorem,.. 44 The Lhuilierian function L,.... 44 The double value of,........ 44 Exercises IX.,.. 45 Breitschneider's analogies,..... 47 CHAPTER III. SOLUTION OF SPHERICAL TRIANGLES. Preliminary observations,..48 Use of auxiliary angles,......48 SECTION I.-The right-angled triangle,.. 49 Six cases of right-angled triangles,.... 49 First case.-Being given c and a, to calculate b, A, B, 49 Type of the calculation,. 50 Exercises X.,... 50 Second case.-Being given c, A, to calculate b, a, B,.. 50 Type of the calculation,........ 51 Exercises XI.,..51 Third case.-Being given a, b, to calculate A, B, c,. 51 Exercises XII.,... 52 Fourth case.-Being given a, A, to find c, b, B,.. 52 Exercises XIII.,.... 53 Fifth case.-Being given a, B, to find b, A,... 53 Exercises XIV.,.... 53 Sixth case.-Being given A, B, to find a, b, c.. 53 Exercises XV.,..54 Solution of quadrantal triangles,... 54 x Contents. PAGE. SECTION II.-Oblique-angled triangles,...... 54 Three pairs of cases of oblique-angled triangles,.. 54 First pair of cases,.... 55 Being given the three sides a, b, c, to calculate the angles,. 55 Type of the calculation,....55 Exercises XVI.,.... 56 Method of calculation when only one angle is required,.. 56 Being given the three angles A, B, C to calculate the sides,. 56 Exercises XVII.,........ 57 Method of calculation when only one side is required,.. 57 Second pair of cases,..58 Being given two sides a, b, and the angle opposite to one of them, to calculate the remaining parts,...... 58 The ambiguous case,..58 Reidt's analogies, advantages of, in calculation,..59 This case solved by dividing the triangle into two right-angled triangles,..... 60 Exercises XVIII.,... 60 Given two angles, A, B, and the side opposite to one of them to solve the triangle,...61 Third pair of cases,......... 61 Being given the sides a, b and the contained angle C to calculate A,,.......... 61 Type of the calculation,....... 61 Exercises XIX.,........ 62 Being given two angles A, B and the adjacent side c to find a, b, C, 63 Cauchy's method of solving the various cases of oblique-angled triangles,.... 63 Exercises XX.,. 65 Brùnnow's series,. 65 CHAPTER IV. VARIOUS APPLICATIONS. SECTION I.-Theory of transversals,... 67 Ratio of section of an arc,........ 67 Anharmonic ratio of four point on an arc of a great circle,. 68 Ratio of section of an angle,.... 68 Anharmonic ratio of four great circles passing through the same point,. 69 Properties of a great circle intersecting the sides of a triangle,. 69 Medians of a spherical triangle,.. 70 The altitudes,........... 70 The common orthocentre of two supplemental triangles,.. 70 Contents. xi PAGE. The Gergonne point of a spherical triangle,.... 70 Anharmonic ratio of a pencil of four great circles,.. 71 Harmonic division of the diagonals of a spherical quadrilateral,. 71 Exercises XXI.,...72 Normal coordinates defined,...... 73 Triangular coordinates,........ 73 Isotomic conjugates,......... 73 Isogonal conjugates,..... 73 Isotomic transversal of a point,....... 73 Isogonal transversal of a point,....... 73 The Lemoine point,. 75 The Symmedian point,.... 75 Normal coordinates of Lemoine point,... 75 Normal coordinates of Symmedian point,..... 75 SECTION II.-Incircles,........ 75 Exercises XXII.,.... 77 SECTION III.-Circumcircles,.... 78 Exercises XXIII.,......... 79 Angles of intersection of the sides of a spherical triangle with its circumcircle,......... 80 Distance between poles of incircle and circumcircle,. 80 Properties of the Lhuilierian,.... 80 SECTION IV.-Spherical mean centres,...... 81 Spherical mean centres defined,..81,, fundamental proposition,... 81 Hart's circle, property of,..82 Exercises XXIV.,.... 84 CHAPTER V. SPHERICAL EXCESS. SECTION I.-Formulas relative to E,..85 Rule of transformation,...85 Cagnoli's theorem,.... 86 Lhuilier's theorem,.... 89 Prouhet's proof of Lhuilier's theorem,.. 89 Exercises XXV.,... 92 SECTION II.-Lexell's theorem,... 92 Steiner's proof,,.... 92 Serret's proof,,....... 93 Steiner's theorem,.... 93 Keogh's theorem,........ 95 Triangle of maximum area, two sides being given,. 95 Exercises XXVI.,.... 97 xii Contents. CIAPTER VI. SMALL CIRCLES ON THE SPHERE. PAGE. SECTION I.-Coaxal circles,... 99 Spherical power of a point with respect to a small circle,. 100 Fundamental property of small circles cutting orthogonally,.100 Coaxal circles defined,........101 Radical circle of the system,...... 101 Limiting points of system,..... 101 Exercise XXVII.,....... 102 SECTION II.-Centres of similitude,. 103 Homothetic points on two circle..... 103 Inverse points on two circles,.......103 Property of centres of similitude of three circles,...104 Homothetic lines on sphere,.......105 Inverse circles....... 105 Exercises XXVIII.,. 105 SECTION III.-Poles and polars,.......106 Harmonic pole and polar defined,......107 Properties of harmonic pole and polar,..108 Exercises XXIX.,.... 110 Anharmonic properties of pole and polar,.. 110 Pascal's theorem,........ 111 Salmon's theorem,.........111 Brianchon's theorem,........111 SECTION IV.-Mutual power of two circles,. 112 Relation between arcs of connection of a point to the vertices of a trirectangular triangle,....112 Distance between two points in terms of the distance from the vertices of a trirectangular triangle,. 112 Mutual power of two circles defined,...113 Frobenius's theorem,....113 Relation between the angles of intersection of any five circles on the sphere,.....114 Condition that four circles should be cut orthogonally by a fifth, 115 Condition that four circles should be tangential to a fifth in terms of their angles of intersection,. 115 The same in terms of their common tangents,. 116 Relation between a system of four great circles and four other circles (great or small),.. 116 Relation between six arcs joining four points on a sphere,. 117 Exercises XXX.,.... 117 Contents. xiii CHAPTER VII. INVERSIONS. PAGE. SECTION I.-Inversions in space,...... 119 Inversions in space defined,......119 The inverse of a plane is a sphere,..... 119 The inverse of a sphere is a sphere,. 120 The inverse of a circle in space is another circle,.. 120 Exercises XXXI.,. 120 SECTION II.-Stereographic projection,.... 121 Stereographic projection defined,...121 Stereographic projection of a circle is a circle,....121 Chasles' theorem,.........121 The projection of the poles of a great circle will be inverse points with respect to its projection,..123 Projection of a system of circles whose planes are parallel,. 123 Circles whose projections are right lines,.... 123 The angle made by any two circles is equal to the angle made by their projections,. 123 Circles 'whose projections are orthogonal to the primitive,..123 The projections of coaxal circles,......123 Application of stereographic projection to spherical trigonometry, 124 Exercises XXXII.,.. 125 Spherical excess of a spherical quadrilateral in terms of its sides and diagonals,.... 126 Spherical excess of a cyclic quadrilateral,. 126 CHAPTER VIII. POLYHEDRA. SECTION I.-Regular polyhedra,.......128 Euler's theorem,..... 128 Proof that there can be but five regular polyhedra,.. 128 Inclination of two adjacent faces of a regular polyhedron,. 129 Polyhedra which are reciprocals,......130 Exercises XXXIII.,.... 131 SECTION II.-Parallelopipeds and tetrahedra,..131 Volume of a parallelopiped,.......131 xiv Contents. PAGE. Volume of a tetrahedron in terms of three conterminous edges and their inclinations,....132 Volume of a tetrahedron in terms of its six edges,...132 The second Staudtians of the solid angles of a tetrahedron are proportional to the areas of the opposite faces,...133 Volume of a tetrahedron in terms of two faces and their included angle,........... 133 Relation between the areas of the four faces of a tetrahedron,.133 Relation between the six dihedral angles of a tetrahedron,. 134 The diagonal of a parallopiped in terms of three conterminous edges,. 134 To find the radius of the circumscribed sphere,..135 First method (Staudt's),..... 135 Second method (Dostor's),.. 136 The isosceles tetrahedron,. 138 Definition of,,....... 138 Properties of,,.... 139 The inscribed sphere touches the faces at the centres of their circumcircles,.........139 Volume of isosceles tetrahedron,.... 140 Radius of circumscribed sphere,.... 140 Radius of inscribed sphere,.... 140 Exercises XXXIV.,. 141 CHAPTER IX. APPLICATIONS OF SPHERICAL TRIGONOMETRY. SECTION I.-Geodesy,....... 143 To reduce an angle to the horizon,. 143 General solution,.... 143 Legendre's solution,. 144 Property of chordal triangle,.... 144 Legendre's theorem,....146 Area of spherical triangle,... 147 Roy's rule,......148 Exercises XXXV.,. 148 SECTION II.-Astronomy,. 149 Astronomical definitions,. 149 To find the time of rising or setting of a known body,. 151 Contents. xv PAGE. Being given the declination and the latitudes, to find the azimuth from the north at rising,...... 151 Being given the hour angle and declination of a star, to find thc azimuth and altitude,... 151 Given the azimuth and altitude, to find the hour angle and declination,. 152 Exercises XXXVI.,.. 152 Given the' sun's longitude, to find his right ascension and declination,...........152 Given the right ascension aud declination of a star, to find its latitude and longitude,.......153 Miscellaneous Exercises,..156 E R R A T A. Page 3, last line, omit "the ".,, 9, line 6, for BD = read BD +., 22,,, 10, for cosaa cos b cos c, read cosa-cosb cosc.,, 72,,, 13, insert a comma after (0 - ABC).,, 112,,, 11,for x, y, z, read sines of x, y, z.,, 131,,, Exercise 4, for three times, read one-third. SPHERICAL TRIGON OMETRY. CHAPTER I. SPHERICAL GEOMETRY. SECTION I.-PRELIMINARE PROPOSITIONS AND DEFINITIONS. 1. DEF. I.-A sphere is the surface generated by the revolution of a semicircle about its diameter, which remains fixed. The term sphere is used in a two-fold signification-1~. As denoting the surface. 2~. The solid bounded by the surface. These correspond to the two-fold signification of the word circle in plane Geometry, namely, the circumference, and the area included within it. DEF. II.-The centre of the generating semicircle is called the CENTIE of the sphere. DEF. III.-A RADItUS of the sphere is any right line drawn from the centre to a point in the surface. DF.,. IV.-A DIAMETER of a sphere is any right line drawn through the centre, and terminated both ways by the surface. From the definition of a spherical surface it follows at once-1~. That every point in it is equally distant from the centre of the generating semicircle. 2~. That any point P in space is outside, on, or inside the surface, according as its distance from the centre is greater than, equal to, or less than the radius. 3~. That spheres having equal radii are equal. B 2 Spherical Geometry. 2. Every section of a sphere made by a plane is a circle. 1~. If the plane passes through the centre, such as ABC, the proposition is evident, since every point in the surface is equally distant from the centre. 2~. When the plane does not pass through the centre, such as DEF. Let O be the centre. From O let fall a perpendicular OIon DEF (Euc. XI. xi.). Take any point F in the section DEF. Join OF, IF. Then, since OI is normal to the plane, the angle OIF is right; therefore IF2 = OF2 - 012; but OF is constant, being the radius of the sphere. Hence IF is constant, and therefore the section DEF is a circle, whose centre is I and radius IF. A l' 0 /F 1 ig 1\p B Fig. 1. Cor. 1.-If R be the radius of the sphere, r the radius of the section, d the distance of the plane of section from the centre of the sphere, r =R2 - d2. (1) Cor. 2.-If R = d, r = 0. Hence the section will reduce to a point, and the plane will touch the sphere. Preliminary Propositions and Definitions. 3 Cor. 3.-Two circles, whose planes are equally distant from the centre, are equal. DEF. V.-The two points P, P' in which the diameter perpendicular to the plane of the circle DEF meets the sphere are called its POLES. From this definition it follows-1~. That all circles whose planes are parallel have the same poles. 2~. That the centre of any circle, its poles, and the centre of the sphere, are collinear. DEF. VI.-A circle of the sphere whose plane passes through the centre is called a GREAT CIRCLE, and a circle whose plane does not pass through the centre is called a sMALL CIRCLE. Thus, on the earth, the meridians, the equator, the ecliptic, are great circles, and the parallels of latitude are small circles. 3. The curve of intersection of two spheres is a circle. A 0/ OJ D Fig. 2. DEIM.-Let any plane passing through the centres 0, O' of both spheres eut them in the circles AEBF, CEDF. Join EF, 00', and produce 00' to meet the circles in A, D. Now 00' bisects EF perpendicularly in I, and it is evident when the semicircles AEB, CED revolve round the line 00' to describe the spheres, that the point E will describe a circle, having tbIfor its centre. B2 4 Spherical Geometry. 4. Either pole of a circle (great or small) on the sphere is equally distant from every point in its circumference. For (see fig., prop. n.), join PF. We have PF2 = PI2 + IF2; but IF2 is constant = R2 - d, and PI2 = (R - d)2. IHence PF is constant. Cor. 1.- PF2 = 2R (R - d). (2) Cor. 2.- P'F2= 2R (R + d). (3) DEF. VII.-A great circle passing through the poles of another circle (great or small) is called a secondary to that circle. DEF. VIII.-The spherical radius of a small circle is the arc of a secondary, intercepted between any point in the circumference and the nearest pole. Thus the spherical radius of the small circle D.EF (see fig., prop. in.) is the arc PD. Cor. 1.-If OA be perpendicular to OP, the point A will describe a great circle. Cor. 2.-The spherical radius of a great circle is a quadrant. This is evident; since P, P' are the poles of the great circle ABC, and AP, CP are quadrants. 5. Only one great circle can be drawn through two points on the surface of the sphere, unless they are diametrically opposite. For only one plane can be drawn through tle centre and the two points, unless they are collinear. Cor. 1.-If two points 4, C be each 90~ distant from a third point P, P is the pole of the great circle, determined by the points A, C. If O be the centre, the line OP is perpendicular to the lines OA, OC, and therefore it is normal to their planes. Hence the line PP' is the axis of the great circle in which the plane 0A C cuts the sphere, and P, P' are its poles. Cor. 2.-If the planes of two great circles be at right angles to each other, their axes are perpendicular, and each passes through the poles of the other. Preliminary Propositions and Definitions. 5 6. The locus of all the points of a sphere which are equidistant from twofixed points A, B of the sphere, is the great circle, which is perpendicular at its middle point to the arc of the great circle AB. DEM.-Let C be the middle point of the chord AB. At C erect a plane P, perpendicular to the chord AB. P passes through the centre of the sphere, and it is the locus of points equally distant from A and B; therefore the points of the sphere, where P intersects it, are the only points on it which are equidistant from the points A, B. Hence the proposition is proved. 7. Any two great circles of the sphere bisect each other. DEir.-Let ABCD, AECF be the great circles; then (Def. vi.) the plane of each passes through the centre of the sphere. Hence the common section A C of these planes passes H G A 0 D Fig. 3. through the centre; but the common section of two planes is a right line (Euc. XI., III.) Hence C is a diameter of the sphere; therefore ABC, AEC are semicircles, and the proposition is proved. 6 Spherical Geometry. DEP. IX.- When two arcs of circles intersect, the angle of the tangents at their points of intersection is called the angle of the arcs. 8. 27he angle of intersection of two great circles is equal to the inclination of their planes. DEM.-Let CG, Cff be tangents to the semicircles ABC, AEC; then (Euc. XI., Def. ix.), since each is perpendicular to OC, the angle between them is equal to the angle of inclination of the planes of the great circles; but (Def. ix.) the angle between CG, CI is the angle of intersection of the great circles. Therefore the angle between two great circles is equal to the inclination of their planes. Cor. 1.-If CB, CE be quadrants, OB, OE are at right angles to OC, and the angle BOE is equal to the angle of inclination of the planes of the great circles. HIence the spherical angle BCE is equal to the angle BOE; but BOE is measured by the arc BE. Hence the spherical angle contained by any two great circles AB C, AEC is measured by the arc of a great circle intercepted between them, and having the point Cfor its pole. Cor. 2.-The spherical angle BAE is equal to the angle BCE. Cor. 3.-The angle of intersection of two great circles is measured by the arc between their poles. For, if BEbe produced, since the plane BOEis perpendicular to OC, BE will pass through the poles of ABC, ÂAEC. Let these be I, K, respectively; then, evidently, the arcs IB, KE are quadrants. Hence IX = BE; but BE (Cor. 1) is the measure of the spherical angle B CE. Hence IK is equal to the measure of the spherical angle. 9. To find the radius of a solid sphere. Sol.-From any point of the spherical surface as pole, with any arbitrary opening of the compass describe a circle ABC; Preliminary Propositions and Definitions. 7 in the circumference of this circle take any three arbitrary points A, B, C, and with the compass transfer the three rectilinear distances AB, BC, CA, and construct a triangle abc on paper, having its sides ab, bc, ca respectively equal to AB, BC, CA. Find i, the circumcentre of the triangle abc. Join ia. P A ~p~~~~c Fig. 4. Erect ip perpendicular to ia, and inflect from a to ip the distance ap equal to the opening of the compass with which the circle ABC was described on the sphere. Erect ap' at right angles to ap, and produce ip to meet it in p'; then pp' is equal to the diameter of the solid sphere. For, from the construction, it is evident, if we join AP', that pp' is equal to PP'. 10. Analogy between the geometry of the sphere and the plane. In order to understand the analogy between plane and spherical geometry, it is necessary to observe that to right lines on the plane correspond on the sphere great circles, and to circles on the plane correspond circles on the sphere, which may be either great or small. On a solid sphere can in general be resolved problems analogous to those on the plane, the instrument employed being the compass. Thus (see fig., ~ 2), if we place one of the points of the compass in P, we can, with an opening equal to the 8 Spherical Geometry. chord PD, describe the circle DEF. To describe a great circle, it is necessary that the opening of the compass should be equal to the chord of a quadrant. We can also, by the compass, divide an angle into 2, 4, 8, &c., equal parts, erect an arc of a great circle perpendicular to another, make a spherical angle equal to a given spherical angle, describe a circle touching three given circles, &c. ExEnCISES.-I. 1. A great circle passing through the poles of two others cuts each at right angles, and their points of intersection are its poles. 2-5. Solve the following problems with the compass:1~. Describe a great circle through two given points of the sphere. 2~. Through a given point of the sphere draw an arc of a great circle perpendicular to a given great circle. 3~. Make, at a given point of a given great circle, an angle equal to a given angle on the same sphere. 4~. Through a given point not on a given great circle draw a great circle making a given angle with it. 6. The loci of the poles of great circles, making a given angle a with a given great circle, consist of two small circles, having the same poles as the given circle. 7. The tangents at a given point A of the sphere to all circles (great or small) passing through A lie in the plane through A, perpendicular to the radius of the sphere drawn to that point. 8. If a tangent line to a sphere passes through a given point, the locus of the point of contact is a small circle. 9. If tangent lines to a sphere be parallel to a given line, the locus of the points of contact is a great circle. 10. The arc of a great circle, perpendicular to the spherical radius of a small circle at its extremity, touches the small circle. 11. Draw a great circle, touching two small circles. 12. Draw a great circle through a given point, touching a given small circle. Spherical Triangles. 9 13. If a variable sphere touch three planes, the locus of its centre is a right line. 14. Describe a sphere, passing through four points which are not coplanar. 15. If A, B, C, D be four points on a great circle, prove that sin BC. sin AD + sin CA. sin BD + sin AB. sin CD = 0. 16. In the same case, prove that sin BC. cos AD + sin CA cos BD + sin AB cos CD = 0. SECTION II.-SPHERICAL TaIANGLES. 11. DEF. X.-The figure formed by the shorter arcs joining three points on the surface of a sphere, no two of which are diametrically opposite, is called a SPHERICAL TRIANGLE. Two points on he surface of a sphere can be joined by two distinct arcs, which together make a great circle. Hence, when the points are not diametrically opposite, these arcs are unequal, and it follows from the definition that each side of a spherical triangle is less than a semicircle. If ABC be the triangle, O the centre of the sphere, the planes OAB, OBC, O CA form a solid angle O - ABC c Fig. 5. (Erc. XI., Def. II.), whose face angles AOB, BOC, COA are measured by the sides of the spherical triangle ABC, 10 Spherical Geometry. and whose dihedral angles (Eue. XI., Def. Il.) are equal to the angles of the spherical triangle (~ 8). There is then a correspondence between the spherical triangle ABC and the solid angle O-ABC: every property of one gives a property of the other. DEF. XI.-The portion of a sphere comprised between two halves of great circles is called a LUNE. Three great circles intersect in six points A, A'; B, B'; C, C'. These are two by two diametrically opposite, and divide the sphere into eight triangles. DEF. XII. —Two triangles BCA, B'CA, which have a common side CA, and whose other sides belong to the same great circles, are called COLUNAR TRIANGLES. The triangle ABC has three colunar triangles, viz. A'BC, B'CA, C'AB. DEF. XIII.-Two triangles ABC, A'B'C', whose corresponding vertices are diametrically opposite, are called ANTIPODAL TRIANGLES. 12. Any two sides of a spherical triangle are together greater than the third, and the sum of the three sides is less than a great circle. DEaM.-Let ABC be the spherical triangle (see fig., ~ 11), O the centre of the sphere. Join OA, OB, OC; then (Euc. XI., xx.) any two of the plane angles forming the trihedral angle O - ABC are together greater than the third; but the arcs AB, BC, CA are the measures of the plane angles A OB, BOC, COA. ience the sum of any two of the arcs AB, BC, CA is greater than the third. Again, the sum of the three plane angles AOB, BOC, COA is (Ecu. XI., xxi.) less than four right angles. ifence the sum of the three arcs AB, B C CCA is less than a great circle. In the same manner it follows that the sum of the sides of any convex spherical polygon is less than a great circle. Spherical Triangles. 11 13. Since every great circle has two poles, it will be necessary to make some convention in order to distinguish them. For this purpose we employ, as in so many other cases, the terms positive and negative. Thus, if BC be an arc of a great circle, its positive pole will be that round which tlie rotation from B to C will be from left to right; that is, in the same direction in which the hands of a watch move, and the other will be the negative pole. For example, if B, C be points on the equator, and C west of B, the north pole will be the positive pole of BC, and the south its negative pole. DEF. XIV.-The spherical triangle, whose angular points are the positive poles of the sides of a triangle ABC is called the POLAR TRIANGLE of AB C. 14. If two spherical triangles ABC, A'B'C' be such that the latter is the polar triangle of the former, the former is the polar triangle of the latter. A C r= \ F _ _ E B Fig. 6. DEM.-Join A'C, B'C by arcs of great circles; then, because A' is the pole of BC, A'C is a quadrant. Similarly, B'C is a quadrant. Hence, since the arcs CA', CB' are quadrants, C is the pole of A'B', and it is evidently the positive pole. Hence the proposition is proved. 12 Spherical Geometry. 15. The sides of either af two polar triangles are the supplements of the angles of the other. DEi.-Produce the arcs A'B', A'C' if necessary to meet BC in E and F. Now, since C is the pole of 'B', the arc EC is 90~. In like manner, the arc BF is 90~. Ience EC + BF or BC+ EFis 180~; but EF is (Art. 8, Cor. 1) the measure of the spherical angle B'A'C'. Hence the side BC is the supplement of the angle B'A' C', and similarly for the other sides and angles. Scholium.-On account of the property proved in this proposition, polar triangles are also called supplemental triangles. Cor.-If we denote the angles of the triangle ABC by A, B, C; their opposite sides by a, b, c; and the corresponding elements in the polar triangle by the same letters accented, we have a'= 180 - A, '= 180 - a. (4) b' = 1800 - B, B' = 180~ - b. (5) ' =180~- C, C' = 180-. (6) 16. In every spherical triangle-1~. The sum of two angles is less than the third increased by 180~. 2~. The sum of the three ngles is greater than two, and less than six right angles. DEM.-1~. From the equations (4)-(6) we get (a + b' - c') = 180 + C- (A + B); but a' + b' is greater than e', therefore 180~ + C is greater than A +B. 2~. From the same equations, we have A + B + C= 540~ - (a' + b' + c'). Hence A + B + C is less than 540~; that is, six right angles. Again (from ~ 12), a' + b' + c' is less than 360~. Hence A + B + C is greater than 180~; that is, two right angles. Spherical Triangles. 13 Cor. 1.-If any side of a spherical triangle ABC be produced, the exterior angle is less than the sum of the two interior non-adjacent angles. Cor. 2.-If all the sides of a convex spherical polygon be produced, the sum of the exterior angles is less than four right angles. DEF. XV.-The amount by which the sum of the three angles of a spherical triangle exceeds two right angles is called the spherical excess. We shall denote it by 2E. Denoting the spherical excess by 2E instead of E has the same advantage as putting 2s for the perimeter of a triangle instead of s, viz., it avoids fractions, and makes certain formulae containing angles symmetrical with the corresponding ones containing sides. Cor. 3.-Any angle of a spherical triangle is greater than E. This is merely another statement of 1~, supra. 17. Two antipodal triangles ABC, A'B'C' are equal in area. DEM.-Two antipodal triangles have evidently equal sides, but are not superposable, except when each is isosceles, because BFig. 7 Fig. 7. their elements are arranged in inverse order. To prove that in the general case the areas are equal. Let P be the pole of the 14 Spherical Geometry. small circle, passing through A, B, C, and P' the pole of the circle through A4'B'C'; then evidently P' is diametrically opposite to P, and the pairs of triangles PAB, P'B'A'; PB C, P' C'B'; PCA, P'A'C', being antipodal and isosceles, are superposable. Hence the area of AB C is equal to the area of A'B' C'. 18. Two spherical triangles on the same sphere have all their corresponding elements equal-l1~. When two sides and the contained angle of one are respectively equal to two sides and the contained angle of the other. 2~. Whien the side and the adjacent angles of one are equal to a side and the adjacent angles of the other. 3~. When the three sides of one are equal to the three sides of the other. 4~. When the three angles of one are equal to the three angles of the other. 'i Fig. 8. Cases 1~, 2~, 3~ correspond to Euc. Book I., Props. Iv., vIII., xxvI. Case 4~ has no analogue in Plane Geometry. It will be sufficient to prove 1~ and 3~, as 2~ and 4~ are inferred from them by the properties of the supplemental triangle. DEM. 1~.-A = A', AB = A'B'; AO = A'C'. If these elements are arranged in the same order, the demonstration follows by superposition, as in Plane Geometry. If they are disposed in an inverse order, such as A'B'C', B C", we can superpose either of them on the antipodal triangle of the other. Spherical Triangles. 15 3~. If the arc A'B' be applied to AB, the point C' will be one of the points of intersection of the arcs of two small circles, described from A and B as poles, and passing through the point C: these arcs will intersect in two points C, C", placed on opposite sides of AB; then, if the elements are disposed in the same order in both triangles, C' will coincide with C. If in a different order, the triangle A'B'C' can be superposed on the antipodal triangle of ABC, and in each case we have the corresponding angles equal each to each. 19. If two sides AB, AC of a spherical triangle be egual1~. The angles B, C are equal. 2~. The median AD, which bisects BC, bisects the angle A. DEM.-The arc AD divides the triangle ABC into two triangles, which are symmetrically equal (18, 3~). Cor.-If two angles B, C are equal, the opposite sides AB, AC, are equal. For the polar triangle of AB C has two sides equal. Ience, &c. 20. Tofind the area of a lune. F E B Fig. 9. Let A CBBD, AEBF be two lunes having equal angles at A; then, by superposition, it is evident that these lunes are equal. 16 Spherical Geometry. Hence, by a process similar to that employed in Esc. VI., I. and xxxiii., it may be proved that lunes are proportional to their angles. Therefore a lune: the whole spherical surface:: angle of lune: 27r. Now if r denote the radius of the sphere, its surface is 47rr2 (Eue., App. 7). Hence, if A denotes the angle of the lune, its area is 2Ar2. (7) 21. Girard's Theorem.Thie area of a spherical triangle = 2-Er2 (Def. xv.). DEM.-Produce the base AB round the sphere, and produce B C, A C to meet it in E and D; also produce CB, CA through B and A to meet again in F. Then the spherical triangle BAF is antipodal to the triangle ED C, and therefore (Art. 17) equal in area to it. Hence the lune C is equal to the sum of Fig. Fig. 0. the two triangles ABC, CED; also the lune A4 = to the sum of the triangles ABC, BCD), and the lune B = to the sum of ABC, CEA. Hence the sum of the three lunes is equal to twice the area of the spherical triangle AB C, together with the area of the hemisphere C = ABGDEIT. Silence, if A denote the area of the triangle AB C, we have 2Ar2 + 2Br2 + 2 Cr2 = 27rr2 + 2A;.. = (A + B +C- C r) r2= 2r. (8) Spherical Triangles. 17 This demonstration is taken from the works of WALLIS, Vol. n., p. 875. The theorem is due to ALBERT GIRARD, a Flemish Mathematician of the 17th century. In 1787, more than 150 years after its discovery, an important application of it was made by General Roy in correcting the spherical angles, observed in the Trigonometrical Survey of Britain, Phil. Trans., Vol. vIII., p. 163, year 1790. See also IMem. Acad., Paris, 1787, p. 358, and Mem. Inst., Vol. vi., p. 511. Cor. 1.-The area of a great circle: area of the spherical triangle:: r: 2E. (9) Cor. 2.-If Z denote the sum of the angles of a spherical polygon of n sides, its area is {( + (2- n)r} r2. (10) EXERCISES.-II. 1. If a triangle coincides with its supplemental triangle, prove that all its sides are quadrants and all its angles right. 2. The sum of two opposite angles of a spherical quadrilateral inscribed in a small circle is equal to the sum of the two others, and each sum is greater than two right angles. 3. The spherical excess of a spherical triangle is equal to the circumference of a great circle diminished by the perimeter of the supplemental triangle. 4. The sum of two opposite sides of a spherical quadrilateral, circumscribed to a small circle, is equal to the sum of the remaining sides. 5. If A, B, C, D be four concyclic points on a sphere, prove that sin ~ BC. s in AD + sin CA. sin i BD + sin AB. sin C CD = 0. (11) This follows from Ptolemy's theorem, since chord BC = 2 sin ~ BC, &c. 6. In the same case, if AB = a, BC=b, CD =c, DA = d, AC=e, BD =f; prove that sin e sin a a. sin d + sin b. sini e (12) sin Sf sin a. sin i b + sin c. sin d() O 18 Spherical Geometry. DEF. XVI.-A spherical triangle ABC is said to be diametrical when its circumcentre is the middle point D of one of its sides AB. This side is called the DIAMETRICAL SIDE. 7. In a diametrical triangle, the angle opposite the diametrical side is equal to the sum of 'the two remaining angles, and is greater than a right angle. 8. Two of the colunar triangles of a diametrical triangle are also diametrical triangles, and the spherical excess of the third colunar triangle is equal to two right angles. 9. If the opposite sides of a spherical quadrilateral be equal, the diagonals bisect each other, and the opposite angles are equal. 10. If in a spherical quadrilateral ABCD the angle A = C and B = D; then the side AB = CD, and BC = AD. Produce the sides AB, CD to meet in E and F; then triangles EBC, FAD have the three angles of one respectively equal to the three angles of the other. DEF. XVII.-If the diagonals of a spherical quadrilateral bisect each other, it is called a SPHERIOAL PARALLELOGRAM. 11. If the four sides of a spherical quadrilateral be equal, the diagonals are perpendicular to each other, and they bisect its angles. Such a figure is called a SPHERICAL LOZENGE. 12. If the four angles of a spherical quadrilateral be equal, the diagonals are equal. 13. In two supplemental triangles ABC, A'B'C', the arcs A', BB', CC' are perpendiculars to the corresponding sides of the two triangles, and the corresponding altitudes of the two triangles are supplemental. 14. The poles of the small circle inscribed in a spherical triangle are also the poles of the small circle circumscribed to its supplemental triangle, and the spherical radii of both circles are complementary. 15. If two small circles on a sphere touch each other, the angle between their planes is equal to the sum or the difference of their spherical radii. 16. The angle of intersection of a great circle and a small circle is greater than the inclination of their planes. 17. The length of a degree on a parallel of latitude is equal to the length of a degree of the equator multiplied by cos lat. For if r be the radius of the equator, and r' the radius of the parallel, then, degree on parallel divided by degree on equator = r'/r = cos lat. CHAPTER II. FORMULAE CONNECTING THE SIDES AND ANGLES OF A SPHERICAL TRIANGLE. 22. A spherical triangle has six elements, namely, the three sides a, b, c, and the three angles A, B, C respectively opposite to them. The triangle is completely determined when any three of the six elements are given, as there exist relations between the given and the sought parts by means of which the latter may be found. The object of this chapter is to establish these relations. Our formulae will be divided into three classes as follows:-The first class includes all formulae into which enter four elements of the triangle. The second those which contain five elements, and the third class the formulae into which enter all six. The formulae which we are going to investigate apply equally to "trihedral angles." The sides of the spherical triangle correspond to the plane angles, forming the trihedral, and the angles of the spherical triangle to the dihedral angles of the trihedral. SECTION I.-FIRST CLASS. 23. There are four Cases of the First Class:I. Three Sides and an Angle. II. Two Sides and the Angle opposite to one of them. III. Two Sides and two Angles, one of which is contained by the sides. IV. Three Angles and a Side. Case I.-Three Sides and an Angle. 24. Let ABC be a spherical triangle, O the centre of the C 2 20 Connecting Sides and Angles of a Spherical Triangle. sphere. Join OA, OB, OC. From any point D in OA draw in the planes AOB, AOC, respectively, the lines DE, DF, at right angles to OA. Then (Euc. XI., Def. ix.), the angle EDF is the inclination of the planes to each other, and therefore (~ 8) is equal to the spherical angle A. Join EF; then, from the plane triangles EOF, EDF, we have EF2 = OE2 + OF2 - 2 OE. OF cos EOF. EF2 = D)E2 + DF2 - 2DE. DF cos EDF. Hence 2 OE. OF. cos EOF = 2 0)2 + 2DE. D. cos A; OD OD) 'F D DE.cos EOF= + - cos A, 0F OE 0E or, cosa = cos b cos + sin b sin c cos A. (13) A B Fig. 11. This is the fundamental formula of Spherical Trigonometry. By interchanging letters we get cos b = cos c cosa + sin c sin a cos B. (14) cos c = cos a cosb + sin a sin b cos C. (15) 25. The formula (13) has been proved only for the case in which the arcs b, c are less than quadrants, to show that they c B<? c A Fig. 12. hold when one of them, c, is greater than a quadrant. Produce First Class. 21 BA, BC to meet in B'. Then, from the triangle B'IAC, in which the sides B'A, A C are less than quadrants, we have cos B' C = cos B'A cos A C + sin B'A sin A C cos B'A C, or cos (r - a) = cos (rr - c) cos b + sin (7r - c) sin b cos (7r - A). Hence cos a = cos b cos c + sin b sin c cos A. If both b and c be greater than quadrants, produce AB, A C to meet in A', and the proposition will evidently hold for the triangle A'BC, and therefore for ABC. 26. By subtracting equation (13) from the identity cos (b - c) = cos b cos c + sin b sin c, we get cos (b - c) - cos a = sin b sin c (1 - cos A). Hence, putting a + b + c = 28, we get sinA s= /sin (s - b) sin (s - c) (16) sin b sin c Similarly, sin 2 B =sin (8s - c) sin (s8 - a) (17) sm c sin a and in C = dsin (s - a) sin (s - b) 8 and sinsin a sin b 27. By subtracting the identity cos (b + c) = cos b cos c - sin b sin c from (13), we get cos a - cos (b + c) = sin b sin c (1 + cos A). Hence cos I- = J s in (s - a) (19) sin b sin c Similarly, cos iB = sin s sin (s -b)(20) sin c sin a and cos C i sin s sin (s-c) ( and cs - sin a sin b (21) The radicals in the formulae (16)-(21) have the positive sign;.-or IA, IB, I C are each less than 90~. 22 Connecting Sides and Angles of a Spherical Triangle. 28. From (16) and (19), we get tan j. = sin ( - b) in (8 - (22) sin 8 sin (s - a) In like manner, tan B = sin (s - C) sin (8 - ) (23) sin s sin (8 - b) and tan C = sin ( - a)in ( - b) (24) sin 8 sin (8 - c) Cor.- tan -A tan B = sin (s- c) (25) sin s tan aB tan C C = 8n ( ) (26) min 8 tan t C tan - sin ( - b) (27) sin s EXERCISES. -III. 1. Prove sin2i* 1 - cos2a - cos2b-cos2c + 2 cos acosb cos 28) 1. Prove sinaAX --- -- - -- (28) sin2 b sin2 c cos etos b cos c b sIbM^ Make use of cos A os b cos c sin b sin c 2.,, cos c = cos (a + b) sin2 7 C + cos (a - b) cos2 O C. (29) 3., cos2 c = cos2 (a + b) sin2 + cos2(a - b) cos C. (30) 4., sin2 C = sin2 (a + b) in2 C + sin2 (a - b) cos2 C. (31) n2 5.,, s in A sin sin C =.sin (32) 2' s in s sin a sin b sine' where n = /sin s sin (s - a) sin (s - b) sin (s - c). (33) The function n is so important in spherical trigonometry that it is convenient to have a definite name for it. PROF. STAUDT*, of the University of ERLANGEN, calls 2n the sine of the trihedral angle 0- ABC (see fig., ~ 24). NEUBERG suggests two other names-1~. The First Stacdtian of the triangle. * CRELLE'S Journal, Band. xxiv., s. 255. First Class. 23 2~. The Norm of the sides of the triangle; and for the function N (see ~ 33) he also suggests the names Second Staudtian, or the Norm of the angles of the triangle. 6. Prove cos A cos B cos = sin.(34) sin a sin b sin c 7., tan A tan i B tan = s —. (35) NOTATIONS.-The arcs which join the vertices of a triangle to the middle points of the opposite sides are called medians, and are denoted by ma, mb, mc, respectively. The arcs of great circles, which are drawn from the vertices at right angles to the opposite sides, are called the altitudes, and represented by ha, hb, h,. The arcs which bisect the interior angles, called the interior bisectors, are denoted by da, db, dc; and the bisectors of the exterior angles by da', db', de'. 8. Prove cos b + cos c = 2 cos osma. (36) 9. Prove in a spherical parallelogram that the sum of the cosines of the sides is equal to four times the product of the cosines of the halves of the diagonals. 10. Prove that the norm of the sides of a triangle is equal to the norm of the sides of any of its colunar triangles. 11. If ABCD be a spherical quadrilateral; and if AB=a, CD = a'; BC= b, DA = b'; AC= c, BD = c', and the arcs joining the middle points of a, a'; of b, b'; and c, c' = a,,, 7, respectively, it is required to prove cos a + b + cos = cosa' os bc cos c' 4os cos y. (37) cos b + cos b' + cos c + cos c' = 4 cos 2 a cos ~ a' cos a. (38) cosc c + cos a + a cos a'= 4 cos b cos b' cos. (39) 12. Prove cos a + cos a' + cos b + cos b' + cos c + cos c' = 2 cos 2 a cos a' cos a + cos b cos 2 b' cos 3 + cos c cos c' cos y}. (40) 13. Prove cosa + os a'+4 cos a cos a' cos a =cos b +cosb '+4 cosb cos b' cos3 = cos c + cos c' + 4 cos c cos i c' cos Y. (41) 24 Connecting Sides and Angles of a Spherical Triangle. 14. Prove cos2 ~a + cos a' cos a cos a' cosa = cos2 b+ cos2 b'+ 2 cos b cos b' cos = cos2 ~ C + cos2 c' + 2 cos c cos a c' cos y. 15. Given the base of a spherical triangle and the sum of the cosines of the sides, find the locus of the vertex. 16. In a spherical quadrilateral the arcs joining the middle points of opposite sides, and the arc joining the middle points of the diagonals, are concurrent. (NEUBERG.) 17. If D be any point in the side BC of a spherical triangle, prove that - cos AD sin BC= cos AB sin DC+ cos AC sin BD. (42) The theorem of this exercise may be called STEWART'S Theorem. It is a generalization of a theorem due to that Geometer.-Sequel to Euclid, Prop. ix., p. 24. 18. If ABC be an arc of a great circle, and AA', BB', CC', arcs perpendicular to any other great circle, prove that sin BC sin AA' + sin CA sin BB' + sin AB sin CC' = 0. (43) 19. Prove n = /(1 - cos2- cob - cosc + 2 cos a cos b cos c). (44) 20. If c be the diametral side of a diametral triangle, prove c~ a b sin2 = sin2 + sin2. (45) 2 2 2 Case II.-Two Sides and the Angles opposite to them. 29. The sines of the sides of a spherical triangle areproportional to the sines of their opposite angles. DEM.-From equations (16), (19) we get, by multiplieation, 2sin i cosi = 2 /sins sin (s - a) sin(s-b) sin (s-c) 2 sin b A cos c A =s sin b sin e or sin = si si n c; (46) a sa 2n sin a sin a sin b sin First Class. 25 sin B 2n In like manner, =: - sin b sin a sin b sin c sin.A sinB sin C Hence. =.-, (47) sin a sin b sin c and the proposition is proved. Or thus:-Let ABC be the triangle, O the centre of the sphere. From any point D in OA draw DG perpendicular to the plane BOC; and from G draw GE, GF at right angles to OB, OC. Join DE, DF. Now since DG is perpendicular to the plane, and GE perpendicular to OB, a line through G parallel to OB would be perpendicular both to DG and GE, A B Fig. 13. and therefore normal to the plane DGE. Hence the angle DEG is equal to the spherical angle B (~ 8). In like manner, DFG is equal to C. Now DE sin DEG = DG = DF sin DFG; therefore DE sin B = DF sin C; but DE = OD) sin D OE = OD sin c, and 1DF= 02 sin DOF= OD sin b. Hence sin B sin c = sin C sin b, or sin B: sin C:: sin b: sin c. 26 Connecting Sides and Angles of a Spherical Triangle. EXERCISES.-IV. 1. If a, b, c be the sides of a spherical triangle, a', b', c' the sides of its supplemental triangle, prove sin a: n: sin c::sin a': sin b': sin c'. (48) 8n3 2. Prove that sinA sinB sin C = s2. si.bin (49) sin2a sm2 b sm2c 3. Prove that tan (A + B): tan (A- B):: tan (a+ b): tan (a-b). (50) 4. Prove that tan (a + a): tan ~ ( - a): tan (B + b): tan (B- b). (51) 5. If the bisector AD of the angle A of a spherical triangle divide the side BC into the segments CD = b', BD = c', prove sin b: sin c:: sin b': sin '. (52) 6. If the bisector of the exterior angle, formed by producing BA through A, meet the base BC in D', and if BD' = c", CD' = b", prove that sin b: sin:: sin b": sin c". (53) 7. Prove that cot A: cot ~ B: cot:: sin (s-a): sin ( - b): sin (s - c). (54) 8. If D be any point in the side BC of a spherical triangle, sin BD sin BAD sin C (55) sin CD = sin CAD sin B' Cor.-If D be the middle point, sin BAD sin B sinb _ - ~~= ~ ~ ~(56) sin CAD sin C sin ce 9. Prove that sin a sin ha = sin b sin hb = sin c sin hc = 2n. (57) 10. Given the base of a spherical triangle and the norm of the sides, prove that the locus of the vertex is a small circle. 11. If mb = m, prove that either b = c, or sin2 a = cos2 b + cos2 ~ c + cos ~ b cos c. 12. If a great circle touch two small circles, whose spherical radii are p, p', and distance between their poles = 8, and if r denote the arc between the points of contact, prove Sin2 rsin2 S - sin2 (p - p') coS p cos p' First Class. 27 13. If AD be the median, prove cot DB =cos a (cosb - cosc) (59) cot2ADB~ 2n () 14. Prove that A A A cot (ina, a) sec a + cot (mb, b) sec b + cot (m, c) sec c = 0. (60) Case III.-Two Sides and two Angles, one of which is contained by the Sides. 30. If we multiply equation (15) by cos b, and substitute the result in (13), we get cos a = cos a cos2 b + sin a sin b cos b cos C + sin b sin c cos A. Hence, transposing cos a cos2 b, substituting sin2 b for 1 - cos2 b, and dividing by sina sinb, we get cot a sin b = cos b cos C +. cos; sin a sin C sin c and substituting, for sinA sin a' we get cot a sin b = cotA4 sin C + cos 0 cos b. (61) This important formula may be enunciated as follows, calling a the first side, and b the second:-The cot of the first, by the sine of the second, is equal to the cot of the angle opposite to the first, by the sine of the contained angle, plus the cos of the contained angle, by the cos of the second side. By interchanging letters in (61), we get cot a sin c = cot A sin B + cos B cos c. (62) cot b sin a = cot B sin C + cos C cos a. (63) cot b sin c = cot B sin A + cos A cos c. (64) cot c sin a = cot C sin B + cos B cos a. (65) cot c sin b = cot C sin 2 + cos A cos b. (66) 28 Connecting Sides and Angles of a Spherical Triangle. EXERCISES.-V. 1. If D be any point of the side BC, prove that cot AB sin DA C + cot A C sin DAB = cot AD sin BA C. (67) 2. cot ABC sin DC + cot A CB. sin BD = cot ADB. sin B. (68) To prove 1, we have (Art. 29), cot c sin AD = cot ) sin j + cos B. cos AD, cot b sin AD = - cot D sin y + cos y cos AD. Eliminate cot D.. To prove 2, we have cot AD sin m = cot B sin D + cos D cos m, cot AD sin n = cot C sin D - cos D cos n. Eliminate cot AD... 3. If we describe a great circle B'D'C', with A as polar, equation (67) gives us tan DD'. sin B'C' = tan BB'. sin D'C' + tan CC'. sin B'D'. (69) A B/'D Fig. 14. a A 4. Prove 2cotma. sin 2 = (cot B + cot C) sin ( a. a). (70) 5. 2 os aot (. ) = ot cot C. (() 6.,, cot ma. tan = os (m. a). (72) 2 sin (B- C) First Class. 29 7. Prove cotb + cot c = 2 cos cot da. (73) A B C 8.,, cos cotda + cos cot db+ cos - cot d, = cot a + cot b + cot c. (74) A B C 9.,, sin - cot da' + sin c cot d + sin cot d' = 0. (75) 2. 2 2( 10.,, cos2a - cos2b = 2n (cos a cot A - cos b cot B). (76) Case IV.-Three Angles and a Side. 31. Multiply (61) by sin a, and (63) by sin b cos c then cos a sin b = sin a cot A sin C + sin a cos b cos C, aînd sin a cos b cos C = sin b cot B sin C cos C + cos a sin b cos2 C. Hence, by substitution and reduction, we get cos a sin b sin C = sin a cot A + sin b cos B cos C. Substitute in this expression for sin a cot A its equal sin b cos A sin B reduce, and we get cos A = - cos B cos C + sin B sin C cos a. (77) Or thus:-Let A'B'C' be the triangle supplemental to ABC; then we have, equation (13), cos a = cos b cos e' + sin b' sin ' cos A'; but a'= r -A, b' = r-B, &c. (Art. 15). Hence, substituting, we get cos A = - cos B cos C + sin B sin C cos a. Interchanging letters in (77), we get cos B = - cos C cos A + sin C sin A cos b. (78) cos C = - cos A cos B + sin A sin B cos c. (79) 30 Connecting Sides and Angles of a Spherical Triangle. 32. If we add (77) to the identity cos (B + C) = cos B cos C - sin B sin C, we get sin B sin C (1 - cos a) = - (cos A + cos (B + C)). Hence (Def. xv.), 2E denoting the spherical excess; that is, A +B+ C-7r, we get sin a = sin E sin (A - E)0) 4 sin B sin C In like manner, sin ib= B-sinE sin(B - E) (81) sin C sin -A and sin i= isin E sin(C -E) (2) sin m4 sinB 33. If we add (77) to the identity cos (B - C) = cos B cos C + sin B sin C), we get sin B sin C (1 + cos a) = cos A + cos (B - C). HIence, cos a = sin(B- E) sin (C-E) (83) sin B sin C In like manner, cosb / sin ( C- E) sin(A - E)(84). sin C sin A and cos c= sin (A -E) sin (B-E) (85) sin A sin B 34. From (80) and (83) we get tan a = \sinE sin(-E) (86) sin (B- ) sin(C - E) ( From (81) and (84) we get tan b = sin Esin (B- E) (87) sin(C - E) sin ( - E) First Class. 31 From (82) and (85) we get tan sinE in (C-) (88) tan C = sin (A - E) sin B - E) Cor. 1.-tan a. tan b = sin E sin(C - E). (89) tan } b. tan c = sin E sin (A - E). (90) tan. t a = sin E sin (B - E). (91) Cor. 2.-tan a. cot 1 b = sin (A - E) sin (B - E). (92) tan b. cot 1 c = sin (B - -E) sin (C- E). (93) tan ic. cot a = sin(C - E) sin (A -E). (94) Cor. 3.tan ja: tan b: tan ic:: sin (A - E): sin(B- E): sin (C-E). (95) ExERacSES. —VI. 1. Prove cosC = - cos (A + B) cos - os (A - B) sin2 c (96) N sin E 2. sin ~ a sin ~ b sin N - (97) 2.",2, si s 9 a =sin A sin B sin C' where N = V sin Esin (A - -E) sin (B - E) sin (0- E). (98) N is called the Norrm of the angles of the triangle. See note, ~ 28, Ex. 5. VN2 3. Prove that cos a cos b cos c = sin s (99) sin E sin A sin B sin C' sin2 E 4.,, tan ltan btanc = s-. (100) 2N 2r 2N 5. sin a = sinC sin b = in -' sin C=. sin B sin GC sin C sin A sin A sin A ' (101) The value N = sin B sin C sin a, and the corresponding value of n, viz.. a sin b sin c sin A, have a remarkable analogy to the equation S =bc sin A in plane trigonometry for the area of a triangle. sin A sin B sin C N 6. Prove si= -a =.-.-. (102) sin a sin b sin c n 32 Connecting Sides and Angles of a Spherical Triangle. 7. Prove 2 sin sinin ha = sinB sin hb = sin C sin h,. (103) 8.,, in a right-angled spherical triangle, having the angle C right, *sin 2E sin inû sil?' sin c = 1 sin-4 sin B' 9.,, in a diametral triangle, having c as the diametral side, sin2e=- cos (nA s B cosc=V cot 4cotB. (104) sin a=V -cotB cot C, cos~a=a/-cosec BcotC. (105) 10. If AD be the bisector of the angle A, prove that cos B + cos C = 2 sin - sin ADB cos AD. (106) cos C- cosB = 2 cos cosDB. (107) cos ~' - cos B = 2 cos - cos ADB. (107) 11. What are the formulae analogous to (106), (107), for the bisector of the external angle? Scholium.-In order to pass from a triangle to its polar, it is useful to remark that we replace a, b, c, 8 s, 4 s, Bs, Cs-, - -, 8-, - E, B-E, C-E, n by wr-A', 7r-B', 7r-C', 7r-E', r-a', r-b', wr-c', A'-E', ' - E', C' -E', s'-a, s'-b', s'- c N. In this manner we could infer the formalae of ~~ 32, 33 from those of ~~ 25, 26. SECTION II.-FIRST CLASS (continued). 35. The Right-angled Triangle. The propositions in ~~ 24-34 connecting four of the six elements of a spherical triangle assume, in the case of the rightangled triangle, a simpler form when one of the four elements is the right angle, some of the terms vanishing, viz., those containing the cosine or the cotangent of the right angle. These modified formulae are obtained as follows, making the angle C right in each: Pirst Class. 33 From the equation sin A sin C sin a sin a sin (~ 29), we get sin A = s. (108) Sm a smc ' ~sin s From the equation cot c sin b = cot C sin A + cos A cos B (~ 30), tan b we get cos =t. (109) tan c From the equation cot a sin b = cot A sin C + cos C cos b (~ 30), tan a we get tanA -. (110) From the equation cos =- cos Ccos + sin C sin cosb (~ 31), we get sin A = b (111) cos b' From the equation cos C= - cos A cos B + sin A sin B cos (~ 31), we get cos c = cot A cot B. (112) From the equation cos c = cos a cos b + sin a sin b cos C (~ 24), we get cos c = cos a cos b. (113) 36. The formulae (108)-(113) may be proved geometrically as follows:-Let ABC be the triangle, C the right angle, 0 the centre of the sphere. From any point D in OA erect D)F at right angles to OA, meeting OC in F, and draw FE at right angles to OC. Join DE. Then FE is perpendicular to FD, because the plane BO C is perpendicular toA O C. Hence DE2 = D.F2 + FE2= O.F2- OD + OE2 - OF2= OE2 - OD; therefore the angle ODE is right. D 34 Connecting Sides and Angles of a Spherical Triangle. FBE FE ED Now OE = D OE; that is, sina = ssinAin c. (108') FD FD BED ' OD EDB ' OD'?,, tanb cosA.tanc. (109') EF EF F-) " -O F^ = D ' OF; tan a= tanA. sinb. (110') OD OD OF OE = OF OE', cos c = cosa cosb. (113/) O.E - OI OFE' OE B A Fig. 15. Multiply together (110') and the formula obtained by interchanging letters, and we get -os aos= tan A tan B. cos a cos b Hence cos c = cot A cot B. (112') Lastly, multiply crosswise (108') and the formula got from (108') by interchange of letters; then sin a cos B tan c = tan a sin A sin c; sin 4 cos c therefore cosB = incos = sin cosb. (111 ) cos a 37. The equations (108)-(111) are easily remembered by comparing them with the corresponding equations for the right-angled plane triangle. Thus First Class. 35 Plane Triangle. Spherical Triangle. a sin a sin A -, sin =, c sin c b tan b cosA = -, cos A =t Ce tan l a tan a tan A = - tan = b >sin b' cos b If in these formulae we interchange the letters A, B, and at the same time the small letters a, b, we get four others, which, however, may be regarded as not essentially different. The formula (113) has a remarkable analogy to Euc. I. xLvI. The formula (112) has no analogue in Plane Trigonometry. 38. Napier's Mnemonic Rules. - If as in the annexed diagram we trace in a plane a pentagon, whose sides have respectively the same numerical measures as the quantities a, 7r A7r 7r a, b, i —,, -B, we obtain a closed figure representing the system of five quanti- - Fig. 16. ties called Napier's Circular Parts. Any one of the five may be selected, and called the middle part, then the two next to it D2 36 Connecting Sides and Angles of a Spherical Triangle. are called the acdacent parts, and the remaining parts the opposites. Thus, if a be selected as the middle part - B, and b are the adjacents, and, and - A the opposites; then 2 2 Napier's rules are sin middle part = product of tangents of adjacents = product of cosines of opposites. These rules will be evident from equation (138)-(113). Though given in most treatises on Spherical Trigonometry, they are disapproved of by some of the ablest writers-Delambre, De Morgan, Serret, Baltzer, and others. We have found by experience that the formulae are easily remembered by the method of ~ 37, which we recommend to the student. EXERCISES.-VII. On the right-angled triangle, Ex. 1-20. 1. Prove that sin2A + sin2 b - sin2 c = sin2 a sin2 b. (114) 2.,, sin2a cos2b = sin (c b) sin(c - b). (115) 3.,, tan2 a: tan2 b: sin2 c - sin2 b sin2 - sin2a. (116) 4.,, cos24. sin2c = sin2e - sin2 a. (117) 5.,, sin2A cos2C = sin2A - sin2 a. (118) 6.,, sin2A cos2 b sinc = sin2c- sin2b. (119) 7.,, os2a cos2B = sin2A - sin2a. (120) 8.,, os2A + cos2 - cos2. COS2 C = Cos2a. (121) 9.,, sin2A - cosB = sin2a sin2B. (122) 10. s 'A si (-b (123) 10.,, sin A= 2cosb sin (13) 11. 2 cos sine (124),, cos A =2 cos b sin ' 12.,, sin(a+b) tan (A + B)=sin (a-b)cot (A - B). (125) cos a + cos b 13.,, sin( +) = os cos(126) 1 + cos a cos b cos b - cos a 14.,, sin(A-B ) (127) 16.,, cos (A + B)=-1 s b (128) sin a sin b 16.,, cos(A/- D) = 1 cos a cos b' (129) First Class. 37 17. Prove that sin2 = sin2 cos cos os22 sin2-. (130) 2 2 2 2 2 A B 18.,, sin(a - b) = sin atan - sin b tan. (131) 2 2 19.,, sin(c- b) = sin (b + c) tan2 -, (132) 20.,, sin (c - a) = cos a sin b tan B. (133) 21. Prove that in an equilateral spherical triangle 2sin A = sec 1 a. (134) 22. If the opposite angles A, C of a spherical quadrilateral ABCD be right, and if the sides AD, BC produced meet in E, prove tanAE. tanDE = tan BE. tan CE. (135) 23. If the internal and external bisectors of the angle A of a spherical triangle meet the base in D, D', prove t sin2b - sin2 a cot DD' = Sm (136) 2 sin a sin b sin c 24. Given the base of a spherical triangle, and the sum of the base angles, prove that the external bisector of the vertical angle passes through a given point. 25. If CC' be the median from the right angle of a right-angled triangle, prove that sin2ina in2b = (2 cos - sin CC')' (137) 26-34. If p be the perpendicular from the right angle C on the hypotenuse, proveC B Fig. 17. 1~. cot2p = cot2a + cot2b. 2~. cos2p = cos2A + cos2B. (138) 3~. tan2a = + tan a' tan c, tan2 b = + tan b' tan c. (139) 4~. tan2a: tan2b:: tana': tanb'. (140) 38 Connecting Sides and Angles of a Spherical Triangle. 5~. sin2p = sin a' sin b'. 6~. sinp sin c = sin a sin b. (142) 7~. tan a tan b = tan c sin p. 8~. tana + tan2 b tan2ccos2p. (144) 9~. cotA: cotB:: sina': sin b'. (145) 35. If MA', MB', MC' be the perpendiculars let fall from a point M on the sides of the triangle ABC; then cos AB'. cos BC'. cos CA' = cos A'B. cos B'C. cos C'A. (STEINER) (146) 36. If the triangles ABC, a8y be such that perpendiculars let fall from A, B, C on the sides of a, j, y be concurrent, the perpendiculars from a, B, y on the sides of ABC are concurrent. 37-40. If AD be the altitude of the triangle ABC, prove1~. cos BD: cos CD:: cos BA: cos CA. (147) 2~. sin BD: sin CD:: cot B: cot C. (148) 3~. tan BD: tan CD tanBAD: tan CAD. (149) 4~. cos BAD: cos CAD: tanBA: tan CA. (150) 41. If the base BC be fixed, and the ratio of the cosines of the sides constant, the locus of A is a great circle perpendicular to BC. 42. If the angle A be fixed, and the ratio tan b: tan c constant, the side BC passes through a fixed point. 43. If the base BC be fixed, and the ratio tan B: tan C constant, the locus of A is a great circle. 44. If the angle A be fixed, and the ratio cos B: cos C constant, the side BC passes through a fixed point. 39. Quadrantal Triangles. The triangle supplemental to a right-angled triangle has one side a quadrant, and is called a quadrantal triangle. The formulae pertaining to such triangles are obtained from the equations (108)-(113) by the substitutions of the Scholium (Art. 33). They are as follows, c being the quadrantal side:sin a = sin 4 - sin C. (151) cos a = - tan B + tan C. (152) tan a = tan A4 sin B. (153) sin a = cos b cos B. (154) cos C = - cot a. cot b. (155) cos C= - cos A. cos B. (156) Second Class. 39 By interchanging the letters a, b, and at the same time the letters, B in the formulae (151)-(154), we get four others, which, however, may be regarded as not differing essentially from those given. SECTION III.-SECOND CLASS. Formulae containing Five Elements. 40. NAPIER'S ANALOGIES.-If we multiply the identity tan A. + tan B tan - (A + B) = tn t- tan i by tan C, 1 - tan -A. tan ^and substitute for tan 4 A. tan 1 C, &c., their values, from ~ 28, Cor., we get tan (A + B) tan sin(s - b) + sin(s - c) cos (a - b) tan } (, + B) tanC —.. G = -- sin s - sin (s - a) cos (a + Hence tan.i(.+B)= ' cotos C. (157) cos ~.(a + b) Similarly, tan (A-B) sin ( cot C. (158) sin n (a + b) Cor. 1.-If in the expression for tan (A + B) we change cos to sin, we get the expression for tan ( (A - B). 41. Again, we have,a,. l tan a a cot c + tan } b cot ce tan - (a + b) cot - e = - tan(a+b)eot1 - tan a a tan b and substituting for tan aa cot 2c, &c., their values from ~ 34, Cors. 1, 2, we get sin(A -E) + sin(B - E) cos (A -B) tan(+b) cot = = sin (C - E) - sin E cos (. + B)' Hence tan (a+ b) = cs ( tan c. (159) cos - (A + B) sin (A - B) Similarly, tan j (a - b) = sitan c. (160) sin 4 (A + B) 40 Connecting Sides and Angles of a Spherical Triangle. Cor. —If we change cos to sin in the expression for tan 2 (a - b) we get that for tan f (a + b). The theorems contained in the equations (157)-(160) may be expressed as proportions, and are called NAPIER'S ANALOGIES, after their discoverer. It may be remarked that the last pair can be got from the first by means of the polar triangle; also that the second and fourth may be inferred from the first and third by multiplying them respectively by the equation tan (A - B) tan 2 (a - b) tan (ZA + B) tan (a + )' Several proofs of these important theorems are known, but the foregoing are probably the simplest. EXERCISES.-VIII. 1. Show that cos a sin b = sin a cos b cos C + cos A sin c. (161) Multiply equation (61) by sin a, and replace sin a cot sin C by cos A sin c. 2. Prove that sin C cos a = cos A sin B + sin A cos B cos C. (162) SECTION IV.-THIRD CLASS. Formulae containing Six Elements. 42. DELAMBRE'S AN&LOGIES.-1~. To prove sin (A + B) cos(a - b) cos 6 C cos c DEM.-sin (A + B) = sin A cos s B + cos A- sin } B; and substituting for sin A, cos }A, &c., their values from ~~ 26, 27, we get sin (+ B) - sin(s-b) +sin(s -a)sins.sin(s-) sin c sin a. sin b Hence sin(A+B) sin(s-b)+sin(s-a) coso(a-b) (16 cos C sin = (163) cos 4 C sin c cos -< Third Class. 41 In like manner we get the three following equations: — 2~. jsin 1 (.A - B) sin (- (a - b) 20~ 2 2 (164) cos 1 C sin c cos (. +B) cos ( a+ b)(165) sin C = cos e 40 ' cos-( - B) sin - (a + b) 40 -1..(166) sin - C sin - ' From any one of these formulae the others may be obtained by the following rule: —Change the sign of the letterB (large or small) on one side of the equation, and write sin for cos and cos for sin on the other. Cor.-Napier's analogies may be inferred from Delambre's by division. Delambre's analogies were discovered by him in 1807, and published in the Connaissance des Temps for 1809, p. 443. They were subsequently discovered independently by GAUSS, and published in his Theoria motus corporis coelestium in sectionibus conicis solem ambientium. Both systems may be proved geometrically, though not so directly as by the method in ~~ 41, 42. The geometrical proof is the one originally given by Delambre. It was rediscovered by Professor Crofton, F. R. S., in 1869, and published in the Proceedings of the London Mathematical Society, Vol. III. 43. REIDT's ANALoGIES. —From Delambre's analogies we get by an easy transformation four others, due to Reidt. See his Sammlung von fAufgaben der Trigonometrie, Seite 233. These may be used in the solution of triangles. Formulae nearly identical with them are given in SERRET'S Trigonometry, p. 156. From (163) we get cos ec- cos C cos s (a - b) - sin ~ (. + B) cos + cos C cos (a - b) + sin (2 + B)' Now, put A + a = 4s', A - a = 4d', B + b = 4s", B - b = 4d", C + c = 4s"', C- c = 4d"', 42 Connecting Sides and Angles of a Spherical Triangle. and we get tan s"' tan d"' = tan (450 - (s" + d')} tan 1450 - (s' + d") }. (167) Similarly, from equations (164)-(166), we get tan (450 + d"') cot (450 + s"') = cot (s' - s") tan(d' - d"). (168) tan (450 + s"') tan (450 + d"') = cot (s' + s") cot (d' + d"). (169) cot s"' tan d"' = tan {450 - (s" - d')} tan {450 - (s' - d")}. (170) 44. From the formulae (167)-(170) we get, by an obvious method, tan2 (45~ - s"') = cot (s' - s") tan (s' + s") tan (d' - d") tan (d' + d"). (171) tan2 (450 - d"')= tan (s' - s") tan (s' + s") cot (d' - d") tan (d' + d"). (172) tan2 s"' = tan i450 - (s" + d')} tan 1450 + (s" - d')) tan (450- (s' +d")) tan [450 + (s'-d")}. (173) tan2 d'" = tan {450 - (s" + d') tan {450 - (s" - d')} tan (450- (8' +d")} tan{450-(s'-d")}. (174) 45. If in the original triangle a, b, B retain their values, and the angle A change into 7r - A, the formulae (171), (172) are replaced by the following:tan2 s"' = tan (s' - s") tan (s' + s") tan (d' - d") cot (d' + d"). (175) tan2 d"' = tan (s' - s") cot (s' + s") tan (d' - d") tan (d' + d"). (176) Third Class. 43 46. Other Applications of Delambre's Formulae. From the 3rd and 4th of Delambre's analogies, we have / c\ A + B c\ A-B co s- cos -sn s- cos2 2 2 C ~ I C\; c ~ IO C\ cos 2 cos 90~ — sin2 cos 90 — 2 2/ 2 2/ From the first of these equations, we get c\ c A + B C\ COS s-) +COS- COS, +Cos 90 — ___.___ 2_ _2 ___ _2 cos - - cos s- - cos 90- - - cos or cot 2cot 2 = cot E tan (C - E); (177) sin s- + sin cos +B cos(90 - 2, 2 2 2 sin(s-2) -sin2 cos 2 - cos (9O-2) 2 2 Hence, by division, and extracting the square root, we have cot s = v/cot -E. tan { (4 - E) tan f (B - E) tan I ( C- E). (179) tan = v/tan E. tan (A - E)tan (B - ) cot }(C- E). 2 (180) These simple and elegant proofs are due to Prouhet. See Nouvelles Annales, 1856, p. 91. 44 Connecting Sides and Angles of a Spherical Triangle. 47. If we put L =,/tanan -. tan ( - ) tan E) tan( C- E), (181) equation (179) may be written 8 L cot 2 tanE' (182) and equation (180) may be written tan s-c L t 2 tan(C-E)' (183) From (183) we get, interchanging letters, tan -tan (-E) (184) s-b L and tan - b=. (185) 2 tan (B-E) 48. If we multiply the four equations (182)-(185) we get L = V/cot 2 s. tan (s - a) tan 2- (s - b) tan 2 (s - c); (186) and substituting this value in (182), we get tan - E = Vtan 1s. tan 2 (s - a) tan 2 (s- b) tan (s - c). (187) This beautiful theorem is due to Simon Lhuilier of Geneva. After him I propose to call the function L the Lhuilierian of the triangle. It will be seen that on account of its double value, viz., those given in (181) and (186), it will give the solution of a spherical triangle either when the three sides or the three angles are given, that the same system of equations solves both cases, and that in each case the solution is selfverifying. Third Class. 45 EXERCISES.-IX. 1. Prove that sin b sin c + cos b cos c cos A = sin B sin C - cos B cos C cos a. (188) 2. Prove Cagnoli's theorem that cos a cos B cos C + cos A cos b cos c = cos A cos B cos C sin b sin c + cosa cosb cosc sinB sin C. (189) tan A tan B tan a 3. Prove tan A tan B tan = + + (. (190) cos b cos c cos b cos c tan a tanb tan c 4.,, tan a tan b tan c =- (191) cos B cos C cos B cos C sin A tan C cos B + cos a sinB 1 " tan b cos C- cos A sin c sin a 6.,, sin C cos b = cos B sin A + cos A sin b cos C. (193) 7.,, sin c cos B = cos b sin a - cos a sin b cos C. (194) tan A sin B - cos b cos c tan a tan b + cos C *8 smin C tan b- tan a cos C( tan a sin b + cos b cos C tanA tanB - cos c 9. = (196) -" sOsin C tanB + tan A cosc 10.,, -- = sin a cos C + cos a cot b. (197) sin B tan c 11. i b tan = sin A cos c - cos cot B. (198) sin b tan 0 s The Exercises 3-11 are due to Barbier. See Nouvelles Annales for 1866, p. 349. The following is an outline of the method of proof:ABC, A'B'C' are two supplemental triangles. To prove 3-A'AI is perpendicular to BC. Hence cos c = cot B cot B, cos b = cot C cot y, A = 3 +y;.*. tan A = tan B + tan y + tan A tan 3 tan y, &c. To prove 1-Compare HK in the triangles A.HK, A'IHK., 5-Apply the formula (61) to the triangles A'HE., AHEK., 6-Compare DC in the triangles DCE, DBC. 46 Connecting Sides and Angles of a Spherical Triangle. tan GAC - tan A To prove 8-tan GAB = tan GAG- tan A 1 + tan GAC.tan A.,, 10-The angles AFC, AFK, are complementary, &c. F E C Fig. 18. The Exercises 4, 7, 9, 11 are inferred from 3, 6, 8, 10 by the properties of supplemental triangles. 12. In a right-angled triangle, prove sin c = N. (199) 13. If the sides of a spherical quadrilateral be a, B, 7, 8, and the diagonals: and (>, prove that A cos (>.) = (cos a cos y - cos 3 cos ô) cosec =. cosec cp. (200) 14. Prove that A cos (ay) = (cos 3 cos 8 - cos:. cos )) cosec a. cosec y. (201) 15. Prove that A cos (85) = (cos a cos y - cos = cos p) cosec / cosec 8. (202) 16. Prove that the angle between the bisector of the angle Cof a spherical triangle ABC and the perpendicular from C on AB =tan-' tan (A + B) tan (A -B) tan. (203) 17.,, (cosA - cos2B) - (cota cosA - cotb cosB) = 2N. (204) 18.,, sin A sin sin in a sin b sin c = 4nN. (205) 19.,, N= 2,2 - sin a sin b sin c. (206) 20.,, N2 = 1 + cos A cos B cos (207) 120 2 cos a cos cos w* 1 - cos a cos b cos ~ Third Class. 47 21. Prove that the sines of the perpendiculars from the orthocentre on the sides of a spherical triangle are proportional to the secants of the opposite angles. 22. If the base BC of a spherical triangle be given in position and magnitude, and the sum of the sides AB, AC be given in magnitude, prove that the locus of the intersection of the bisector of the external vertical with a great circle perpendicular to AB at B is a great circle. 23-30. Prove the following analogies due to Breitschneider. See CRELLE'S Journal, Band. xiii., Seite 145:sin E. cos (A - E) sin s. sin (s- a) I ~. n --- -. - 2 (208) sin 3 A cos ~ a cos 2 E. sin (A - E) cos s. cos (s-a)2. 2- - ---— 2 ---- - OS — 2__,=. (209) sin 2 A cos a sin E. sin (A-E) sin(s-b) sin (s -c) 30. -_= 2 — - --------. 2 2U (210) cos 2A cos a cos E. cos (A-E) C cos ( s-b c)c (21) cos A cosa a cos (B- E) cos E) sin s.cos ~ (s - a) 50.. = 2.-2 __ — _-. (212) sin A sin ~ a sin' (B-E) sin~(C-E) cos~s.sin (s-a)213 60. -f -- _ _ (213> sin A sin a 7 cos(B — E) sin (C-E) sin(s-b) cos(s-) (214) 7. 2 — --- - - -- - - - -- - - -- 2 --------; -- _ ----------, ~(2 14 ) cos ~A sin a 8. sin(B-E) cos(G-BE) cos~(s-b) sin(s-c) COS 2 Z 51n 2 a cosS A sin 1a(215 These analogies are all inferences from Delambre's. For example, 1~ is obtained by subtracting both sides of the equation cos ~ (B + C) cos ~ (b + c) n -c from unity. sin 'A cos a CHAPTER III. SOLUTION OF SPHERICAL TRIANGLES. 49. Preliminary Observations. 1~. The logarithms of trigonometrical functions are obtained from their "Tabular Logarithms " by subtracting 10 from the characteristics. For example, log tan 37~ 40' 16" = 1. 8876649. The ablest recent continental writers, such as SERRET, BRIOT, et BOUQUET, and others, employ the logarithms thus reduced, instead of the Tabular Logarithms. We may add that the late PROF. BOOLE was strongly in favour of this alteration. 2~. It is necessary to avoid the calculation of very small angles by their cosines, or of angles near 90~ by their sines, for their tabular differences vary too slowly. It is better to determine such angles, for example, by means of their tangents. 3~. When angles greater than 90~ occur in calculation, we replace them by their supplements, and if the functions of such angles be either cos, sec, tan or cot, we take account of the change of sign. 4~. Formulae not adapted to logarithmic computation can be rendered so by means of an auxiliary angle. Thus:(a) For A cos a + B sin a we put A = B tan E, which gives B sin (+ a) (216) cos 4 (b) For A cos a + B we put B = A sin a tan 4, whic -is.A cos(a-<) which gives cos - (217) cos 9 The Right-angled Triangle. 49 SECTION I.-THEr RGIHT-ANGLED TBIANGLE. 50. The solution of right-angled spherical triangles presents six distinct cases, which correspond to and are solved by the six equations (108)-(113) of ~ 35. For their discussions the following remarks are useful:-1~. The three sides of a spherical triangle (omitting triangles birectangular or trirectangular), are either ail acute, or else one is acute and the other two obtuse. This follows from the equation cos c = cos a cos b. 2~. Either of the sides containing the right angle is of the same species as the opposite angle. This can be inferred from the equation cos A = cos a sin B. It will be a useful exercise for the student to prove these propositions geometrically. 51. FIRST CASE.-Being given c and a, to calculate b, A, B. The required parts are given by the formulae COS ~ cos b= -, equation (113). (218) cos a sin a sin in c' (108). (219) tan a cosB= tan e' (109). (220) The formula (219) gives two supplementary values of sin A, but the ambiguity is removed by considering that A must be of the same species as a. From the equations (218)-(220) we get, by obvious transformations, the three following:tan i b = + V/tan - (c + a) tan - (c - a). (221) tan (45 + )A) = I /t-n (C- ) (222) au-lanec - a( ) tan B=+ in (c - a) (223) sin (c + a) E 50 Solution of Spherical Triangles. The equation (221) proves that if 2 (c + a) be greater than 90~, c is less than a, for the product of the quantities under the radical must be +. The sign is + or - in (222), according as a is less or greater than 90~. If the given parts c and a be each 90~, the angle A is 90~, and b is indeterminate. It is also evident, from the formulae (218)-(220), that c must lie between a and -r - a, in order that the values of cos b, cos B, and sin A may be numerically less than unity. EXAMPLEGiven c = 37~ 40' 20", a = 37~ 40' 12"; find b, A, B. Type of the Calculation. c + a = 75~ 20' 32", (C + a) = 37~ 40' 16". c-a= O 0 8, i(c-a)= 0 0 4. 1 tan (c + a) = 1'8876649, I sin (c + a) = 1i9856305, I tan i (c - a) = 5-2876348; 1 sin (c - a) = 5-5886648;.. I tan } b = 3-5876498..~. I tan (45 + i A) = 2-3000150, I tan aB- 3-8015172. Hence b = 0~ 26' 37"'2, A = 89~ 25' 37", B = 0 43' 33". EXERCISES.-X. 1. Given e = 63~ 55' 43", a = 120~ 10' 0"; find b,, B. 2., c=54 20 0, a= 36 27 0;,, 3.,, c=87 11 39-8, a= 86 40 0;,, 52. SECOND CAsE.-Being given c, A, to calculate a, b, B. The unknown parts are found thus:sin a = sin c sin A, equation (108). (224) tan b = tan c cos A,,, (109). (225) cot B= cos c tanA,,, (112). (226) The Right.angled Triangle. 51 The ambiguity in finding a by its sine is removed in ~ 51. If a be very near 900, we commence by calculating the values of b and B, and then determine a by either of the formulae tan a = sin b tan A. (227) tan a = tan c cos B. (228) EXAMPLEGiven c = 81~ 29' 32", A = 32~ 28' 17"; find a, b, B. Type of the Calculation. c = 81~ 29' 32", A = 32~ 28' 17". 1 sin c = 1-9951945 1 tan c = '8250982 sin. = 1-7278843 1 cosA -= 1*9269687 I sin a ^ 1-7230788; 1 tan b = -7520669;. a = 31~ 54' 25"..'. b = 79~ 51' 48"'65. I cos c = 1*1700960 I tan`A= 1.8009157 Icot B = 2-9710117;. B = 84~ 39' 21"-33. EXERCIES. -XI. 1. Given c= 69~ 25' 11", A = 54~ 54' 42"; find a, b, B. 2.,, c=112 48 0, A= 56 11 56;,,,, 3.,, c= 46 40 12, 4=37 46 9;,, 53. THIRD CASE.-Being given a, b, tofind A, B, c. The formulae aretan A = tan a sin b, equation (110). (229) tanB =tan b sin a,,, (110). (230) cos c = cos a cos b,,, (113). (231) E2 52 Solution of Spherical Triangles. If the side c be very small, instead of the formulae (231), we may determine it by means of either of the following equationstan c = tan b - cos A = tan a - cos B, (232) A, B having been calculated from the formulae (229), (230). EXERCISES.-XII. 1. Given a = 120~ 10' 0", b = 150 59' 44"; find A, B, c. 2.,, a= 36 27 0, b= 43 32 31;,,, 3.,, a= 86 40 0, b= 32 40 0;,,,, 54. FOUIRT CASE.-Being given a, A, tofind c, b, B. In this case we have sin c = sin a ' sin, equation (108). (233) sin b = tan a - tanA,, (110). (234) sinB = cos A- cos a,,, (111). (235) Since each of the sought parts is found by its sine, there will be on the whole six solutions, the formulae (233)-(235) giving two values for each of the sought parts c, b, B. The parts a, A must be of the same species (see ~ 50). Also sin a must be less than sin A (A is comprised between a and 90), and the formula (233) gives two admissible values of c, say c, and 180 - cl; the formula (234) gives two values of b, b, and 180 - b,, one of which goes with c,, and the other with 180 - c,; for the three sides a, b, c are all acute, or one alone is acute; the formula (235) gives two values of B, of which one goes with b,, and the other with 180 - bi (b and B are of the same species). If the sought parts are badly determined by the equations (233)-(235), which happens when they are near 90~, the preceding formulae may be written as follows (see Notation, ~ 43): The Right-angled Triangle. 53 tan (45~ + e) + tan 2gs (236) tann 2' tan (45 + b)= + sin 4s' sin 4d'r (287) tan (45~ + )= + |ot 2g. (238) cot 2d' Each of the radicals on the right-hand side must have the double sign. EXERCISES.-XIII. 1. Given a = 34~ 6' 13", A=34~ 7' 41"; findc, b, B. 2.,, a=87 12 28, A=87 51 37;,, 55. FIFTH CASE.-Being given a, B, tofind b, A, c. The following equations give the required partstan b = sin a tan B, equation (110). (239) cos A = cos a sin B,,, (111). (240) tan c = tan a -cosB,, (109). (241) If A be small instead of the equation (240), the following may be usedtan a tanA =- ta-, equation (110). (242) sin b EXERCISES.-XIV. 1. Given a = 92~ 47' 32", B=50~ 2' 1"; findb, A, c. 2.,, a=96 49 59, B=50 12 4;,, 3.,, a=20 20 20, B=38 10 10;,, 56. SIXTH CASE.-Being given A, B, tofind a, b, c. Here we have cos a = cos A sinB, equation (111). (243) cos b= cos B sin A,, (111). (244) cos = cot. cotB,,, (112). (245) 54 Solution of Spherical Triangles. If these formulae be not well adapted for the arithmetical calculation, which happens when the sought parts are small, we use the followingtan a =+ tan ( - E) (246) tan b=+ = ttata -E (247) tan (A - E) tan c = + - (248) tanos (A-) (248) EXERCISES.-XV. 1. Given A = 63~ 15' 12", B = 163~ 33' 39"; find a b, c. 2.,, A=46 59 42, B = 57 59 17;,,, 3.,, = 421 24 9, B= 99 4 11;,,,, 57. The triangle supplemental to a right-angled triangle is a quadrantal spherical triangle, that is, a triangle one of whose sides is a quadrant. The solution can be inferred from the equations of ~ 39. Other triangles besides the quadrantal can be reduced to the rectangular. 1~. Isosceles triangles, for the median that bisects the base is perpendicular to it. 2~. Triangles in which a + b = 7r, or A + B = 7r, for the colunar triangle B'A C is isosceles. SECTION II.-OBLIQUE-ANGLED TRIANGLES. 58. The solution of oblique-angled triangles presents three pairs of cases, each consisting of two which are reciprocals of each other. They are as follows:I.-The three sides and its reciprocal the three angles. II.-Two sides and the angle opposite to one of them; its reciprocal two angles and the side opposite to one of them. III.-Two sides and their included angle; its reciprocal two angles and the adjacent side. Oblique-angled Triangles. 55 First Pair of Cases. 59. Being given the three sides a, b, c, to calculate the angles. From equation (186) and equations (182)-(185), we get logL-= fI cot s+ltan- (s-a)+ltan -(s-b) +tani (s- c). (249) ItanLE=log - cot s. (250) tan 2 (A - E) = log L - 1 tan I (s-a). (251) tan i (B - E) = log L - tan i (s - b). (252) i tan } (C - E) = log L - 1 tan - (s - c). (253) EXAMPLEGiven a = 100~, b = 37~ 18', c = 62~ 46'; find A, C. Type of the Calculation. a= 100~ 0 0" cotes = 1-9235570 b= 37 18 0 tan (s -a) = 4-4637261 c= 62 46 0 tan (s -b)= 17150481 8s= 50 1 0 tan (s - c) = 1-5278682 i-(s-a)= O 1 2 L5-7001994 -(s-b)= 31 22 0.'. logL = 3-8500997 -(s-c)= 18 38 0 Hence tan -E = 39265427,, tani(Â-.E) = 1-3863736,, Titan (B-E) = 2-0650516,, Itan~(C-.E)= 2-3222315.. E= 0~ 58' 3"32, A = 176~ 15' 46"'56, B=2~ 17' 55"'08, C = 3 22' 25"'46. 56 Solution of Spherical Triangles. EXERCISES.-XVI. 1. Given a= 89~ 59' 59', b= 88~ 55' 58", c= 87~ 57' 57"; findA,B, C. 2.,, a=120 55 35, b=59 4 25, c=106 10 22;,, 3.,, a= 20 16 38, b=56 19 40, c= 66 20 44;,, 60. If only one angle, say A, be required, the formula for tan A in terms of the sides is simpler than the foregoing. Thus in logarithms, I tan ~ A = ~ { sin (s - b) +1 sin (s - c) - I sins - l sin (s - a)}. (254) a = 82~ 33' 51" 1 sin (s - b) = 1*9788195 b =27 16 9 sin(s - c) = 12529286 c= 89 12 24 1-2317481 s =99 31 12 sins = 1-9939773 s-a=16 57 21 1 sin(s-a)= 1-4648388 - b = 72 15 3 1-4588161 s-c= 10 18 48 [.l7729320 l. tan I A = 1-8864660 Hence A = 75~ 11' 22". 61. Being given the three angles A, B, C, to calculate the sides. From equations (181)-(185) we have logL=: { tan E+ 1 tan (A-E)+ Itani(B-E)+ltan~( C-E)}. (255) l cot l s = log L - 1 tan E. (250') I tan (s - a) = log L - l tan (.A - E). (251') 1 tan ~ (s - b) = log L - 1 tan } (B - E). (252') 1 tan i (s - c) = log L - 1 tan - (C - E). (253') On comparing the equations (250')-(253') with (250)-(253) of ~ 59, it will be seen that they are identical Oblique-angled Triangles. 57 EXERCISES.-XVII. 1. Given A = 161~ 22' 10", B = 26~ 58' 46", C= 39~ 45' 10"; find a, b, c. 2.,, A= 127 22 7, B= 128 41 49, C= 107 33 20; find a, b, c. 3., A= 78 15 41, B=153 17 6, C= 87 43 36; find a, b, c. 62. If only one side, say a, be required, we may use either of the formulae tan 1= e sinE.sin(-iE) = sin(B -E)sin (C -E) cos A + cos B cos C or cos a = sin B sin C The last can be adapted to logarithmic computation by means of an auxiliary angle. Thus, if we put sin B cos C tan 4 = cos A sin(B + >) cot C we get cosa= sinBsin ' (256) sin B sin (b EXAMPLE. -Given A = 320 54' 28", B = 146 58' 9", C= 24~ 54' 47"; find a. I sin B = 1-7364682 I cot C = 0-3330492 I cos C = 1-9575824 I sin (B + =) = 2-6464053 1-6940506 2'9794545 1 cosA = 1-9240447 l sin B = 1-7364682 i tan q = 1-7700059 I sin 4 = 1-7053630 Hence b = 30~ 29' 30"'4 1-4418312.B + + = 177 27 39-4..'. I cos a = 1-5376233. Hence a= 69~ 49' 40". 58 Solution ofSpherical Triangles. Second Pair of Cases. 63. Given two sides a, b, and the angle A opposite to one of them, to calculate the remaining parts. The sought parts are found by the following equations: sin A. sin b sin B== (257) sin a tan = tan (a - b) si( ) (258) sin (A -B)' tan x C= cot (A-B) sin (a - b) (259) sin (a + b)' The formula (257) gives for B two values BI, 180~ - B1, if sin A sin b be less than sin a. In order that either of these may be admissible, it is necessary and sufficient that, when substituted in (258), (259), they give positive values for tan 2 c and tan - C, or, which is the same thing, that a - b and A - B will be of the same sign. This condition is both necessary and sufficient. For a, b, A, being the given elements, denote by B, c, C the other elements determined by the equations (257)(259). Now let us construct a triangle T, having the angle C and the sides a, b, and calling A', B', c' the other elements of this triangle, we have tan C'= tan (a - b) s-n (a' + B') (258) t 2 smn ( A ' (25) B) tan 'tan B) sin 2- ('- B')' tan i C= cot(A-B') sin - ( - b) ( ) sin a (a + b)' tan (.A' + 2') tan (a + b) tan (A' - B') tan - (a - b)' and from (257) we infer tan- ( A+ B) tan 2 (a + b) tan (A - B) tan (a - b)' Oblique-angled Triangles. 59 If we compare (259) and (259'), we see that A'- B'= A - B. Hlence, from the two formulae (a) we have À' + B' = A + B; therefore A'= A, B' = B. Lastly, from (258) and (258') we infer that c' = c. Hence we have the following Rule:-If each of the two values of B which are got from (257) be such as that (A - B) and (a - b) have like signs, there are two solutions. If only one of them satisfies this condition, there is only one triangle that satisfies the problem. The problem is impossible when neither of the values of B make (A - B) and (a - b) of the same sign. Instead of the formulae (258), (259), we may use the following: tan c = tan (a + b) + 1 cos 2(A + B) - I cos ] (A - B). (260) Titan ' C = 1 tan (A. + B) + L cos (a - b) - l cos (a + b). (261) 64. From REIDT'S Analogies (~ 44) we get the following equations:1 tan (45~ - d"') = L { tan (s' + s") + I tan (s' - s") + l tan (d' + d") - tan (d' - d")}. (262) I tan (45~ - s"') = 1- { tan (s' + s") - 1 tan (s' - s") + l tan (d' + d") + tan (d'- d")}. (263) These formulae determine C and c when the angle B is acute. They possess the advantage of requiring only four logarithms instead of six, which are necessary if we calculate by the equations (258), (259). For the second triangle answering the given conditions, or for B obtuse, the formulae areI tan s"' - {I tan (s' + s") + 1 tan (s' - s") + I tan (d' - d") - tan (d' + d")}. (264) I tan d"' + ( {I tan (s' - s") + 1 tan (d' - d") + 1 tan (d' + d") - Titan (8' + 8")}. (265) 60 Solution of Spherical Triangles. The formulae (263)-(265) may be replaced by the following:I tan (45~ s"') = I tan (s' + s") + tan (d' + d") - tan (45~ - d"'). (266) I tan d' = I tan (s' - s") + I tan (d' - d") - 1 tan s"'. (267) Or thus:-Let fall the perpendicular CD; then, denoting the arc AD by 4 and the angle A CD by I, we have, from the right-angled triangle A CD, tan 4 = tan b cos A, tan ti = cot A/cos b. C D Fig. 19. Then the sought parts are given by the equations sin B = sin b sin A/sin a, cos (c - )) = cos a cos 4/cos b, cos (C - f) = cot a tan b cos i. EXERCISES.-XVIII. 1. Given a= 73~ 39'38", b=120~ 55' 35", A = 88~ 52' 42"; find B, C, c. 2., a=150 57 5, b=134 15 54, A=144 22 42; find B, C. c. 3., a= 20 16 38, b= 56 19 40, A= 20 9 54; find B, C,. Oblique-angled Triangles. 61 65. Given two angles A, B, and the side a opposite to one of them, to solve the triangle. The solution may be inferred at once from the reciprocal case, ~~ 63, 64. In fact, the same equations solve both cases. Third Pair of Cases. 66. Being given the sides a, b, and the contained angle C, to find, B, c. Napier's analogies give tan I (A + B) = 1 cot} C+ 1 cos 1 (a - b) - I cos 1 (a + b). (268) I tan } (A - B) = I cot - C + I sin - (a - b) - 1 sin 1 (a + b). (269) These equations give 1 (A + B) and 1 (A - B); and therefore A and B, and then c, can be found from equation (47), or from (159) or (160). EXAMPLE.-Given a = 113 2 56", b = 82~ 39' 28", C= 1380 50'14"; find, B, c. Type of the Calculation. 3(a-b) = 15~ 11' 14" 1 sin (a - b) = 1-4184891 -(a+b)=97 51 12 sina-(a + b) = 1-9959075 6 C = 69 25 7 1 cos 1 (a - b) = 1-9845438 {- cos (a+ b) = 1-1355722 cot 3 C = 1-5746163 Hence l{-tan- (A+B)} = 4235869.. I(A + B) = 110~ 39' 35", and l tan 1 (A - B) = 2-9971969; - (A - B) = 5~ 40' 27". Hence A = 116~ 20' 2", B= 104~ 59' 8". 62 Solution of Spherical Triangles. To find C we have, from (159), Itan - c = {-tan -(a+ b)} + [-cos(A+ B)} -1 cos(A-B). Now - tan 2 (a + b)} = 8603353, {- cos (2 + B)} = 1-5475498, I cos - (A - B) = 1i9978668. Hence tan -c =-4100083;.. c= 137~ 29' 3". Observation.-In the foregoing calculation it is seen that, when an angle is between 90~ and 180~, we have the sign minus before its cosine and its tangent, the reason of which is obvious. Or thus:-Let fall the perpendicular BE; then denoting the arc AE by 0, CE will be b - 0. Then, from the right-angled triangles, we have tan (b - 0) = tan a. cos C, tan A/tan C = sin (b - 0)/sin 0. B -4 Fig. 20. The first equation determines 0, and the second A. In a similar manner B may be found. Lastly, from the same triangles, we have cos c/cos a = cos 0/cos (b - 0). Hence c is found. EXERCISES.-XIX. 1. Given a= 88~ 12'20", b = 124~ 7'17", C =50 2' 1"; find A, B, c. 2.,, a=110 55 35, b= 88 12 20, C=47 42 1; find A, B, c. 3., a= 65 15 12, b= 47 42 1, C= 59 4 25; find, B, c. Oblique-angled riangles. 63 67. Given two angles A, B, and the adjacent side c, to finJ a, b, C. From Napier's Analogies we get I tan 1 (a + b) = 1 tan- c + 1 cos (L(A - B) - i cos 2- (A + B). (270) tan 1 (a - b) = I tan 1 c + 1 sin (A B - B) - 1 sin (A. + B). (271) Hence a, b are known, and C can be found from (268) or (269). Or thus:-Let fall the perpendicular BE (see last fig.); then denoting the angle ABD by <, the angle DBC will be B -. Then from the triangles ABD, CBD, we get cot b = tan A cos c, tan a = cos s tan C/cos (B - ). The first of these formulae determines <, and the second a. Similarly b may be found. Again, from the same triangles, we have sin 4: sin (B - ):: cos: cos C. Hence C is found. 68. The following simple and elementary methods of solving the various cases of oblique-angled triangles, by dividing each into two right-angled triangles, are due to CAUCHY. _ c \ 1/ D Fig. 21. Let 0 be the centre of the sphere, AB C the spherical triangle. 64 Solution of Spherical Triangles. Draw CD perpendicular to the plane A OB, also DA', DB' perpendicular to OA, OB. Then it is evident that CA', CB' are also perpendicular to OA, OB. Let B'OD = a, DOA'= 3, DOC= 8; then we have CB'=sina, OB'= cos a; CA'=sinb, OA'=cos b; CD = sin 8, OD = cos 8; CAD=' - A,*CB'D = B. The triangles ' OD, B'OD, A' CD, B' CD, have the angles A' B' D right; OB' OA' DO = cos - = co Bs3OD cosA' OD' cos a cos b Hence s (272) cos o coQ B' DC = sin 8 = CB' sin C7'D>= CA'.sin CA'D. Hence sin a. sin B = sin b sin A. (273) DB' = OB' tan B'OD = CB' cos CB'D. Hence tan a = tan a cos B. (274) D.' = OA' tanA'OD = CA' cos CA'D. Hence tan / = tan b cos 4. (275) From (272) we get tan 2 (a + b) tan ( (a - b) = tan - (a + /) tan - (a -,); but (a + /3) = c. (276) Hence tan 1 (a + b) tan i (a - b) = tan i G. tan } (a - f). (277) The formulae (272)-(277) solve all the cases of obliqueangled triangles. Ist Case.-Given the three sides. (276) gives (a + /), (277) (a - 3), (274), (275) give A, B. 2nd Case.-Given the sides a, b, and the angle A. (273) givesB, (274), (275) give a, /, (276) gives c, and (273) C. 3rd Case.-Given the angle A and the adjacent sides. (275) gives /, (276) a, (272) a, and (273) determinesB and C. Oblique-angled Triangles. 65 EXERCISES.-XX. 1. Prove that cos (a + b) cos (a - b) tan C = sin a cos B + cos sin b. (278) 2-7. Solve a right-angled triangle, being given-1~. c, a+ b; 2~. c, a-b; 3~. a, b+c; 4~. a, b-c; 5~. c, A-B; 6~. c, p. 8. If 1 t cos a + cos b + cos c = 0, prove that each median is the supplement of the corresponding side. 9. In the same case, prove that the spherical excess is two right angles. 10. In the same case, prove that the arcs joining the middle points of two sides are each = 90~. 11-17. If a + b + c = ir, proveA 1~. cos a = tan 1 B tan ~ C. 2. sin2 = cot b cot c. 3. cos2 A 4. tan2 a A cos b cos c 2 sin b sin c 2 cos a A B C2 5~. sin2- + sin2- + sin2 = 1. 6. cos.A + cos B + cos = 1. 2 2 2 7~. cosec (A - E) + cosec (B - E) + cosec (C - E) = cosec E. 18. ABC is a spherical triangle right-angled at C; if with A, B as poles great circles LHFKL, DEFG be describcd, meeting the sides CA, CB, AB of the triangle in the pairs of points E, H; K, G; L, D, respectively; prove that the five triangles ABC, ADE, HEF, FGK, KLB have all the same circular parts.-(NAPIER.) 19-22.-Deduce from the analogies of DELAMBRE or NAPIER the following convergent series:a-b c A B A B 1~. = - cot - tan - sin c + 1 cot2 -tan2 sin 2c - &c. 2 2 2 2 2 2 2 (BRUNNOW). (279) 2~. - + cot- tan sin (a - b) + cot2 - tan2- sin 2(a- b)+&c. 2 \2/ c 2 2 (Ibid.) (280) a+b c A B A B 30. = tan - tan sin c + tan2 tan2 - sin 2e + &c. 2 2 2 2 2 (Ibid.) (281) c +b. B+b 4~. -tan tan sin (a+ b) +tan tan2 sin 2 ( + b)-c. 2 22 2 2 (Ibid.) (282) F 66 Solution of Spherical Triangles. 23. Prove cos 2A + cos 2b - (cos 2a + cos 2B) = cos 2A cos 2b - cos 2a cos 2B. (283) 24. Given any three of the six quantities s', s", s"', d', d", d"' of ~ 43, solve the triangle. 25-32. Solve a spherical triangle, being given-1~. A, B, a+ b; 2~. A, B, a- b; 3~. C,c, a+b; 4~. a,B, C6+c; 5~. c, b, h; 6~. a,B,E; 7~., b, a c; 8~. ab, E, &c. 33. Prove tan (45~ - s') cot (45~ = d') = cot (s" - s"') tan (d" - d"'). (284) 34. Prove tan (45~ - s') tan (45~ -'d') = tan (s" + s"') tan (d" - d"'). (285) 35. Prove tan (B + C) + tan (B - C) = 2 cot sin b - sin (b + c). (286) 36. If the cosines of the tangents drawn from any point P to two small circles have a given ratio, prove that the locus of P is a great circle. 37. In the same case, if the sum or the difference of the cosines of the tangents be given, prove that the locus of P is a circle. 38. State and prove the series similar to (279), (280) that may be obtained from the first and second of Napier's Analogies. 39. If ci, c2 be the values of the third side, when A, a, b are given, and the triangle is ambiguous, prove that tan tan = tan (a + b) tan (a-b). (287) 2 2 40. If A, B, C, &c., be the angular points of a regular polygon of n sides inscribed in a small circle, whose spherical centre is 0 and radius r; prove, if P be any point on the sphere, that Z cos AP = n cos r. cos OP. (288) CHAPTER IV. VARIOUS APPLICATIONS. SECTION I.-THEORY OF TRANSVERSALS. 69. DEF. XVIII.-Being given three points A, B, X on the sin XA same arc of a great circle, the ratio i X is called the ratio sin XB of section (AB, X). The arcs XA, XB are considered of the same or of different signs, according as they are measured in the same or in different directions from X. It is seen that it makes no difference whether we take for XB the arc - XMIB or + XAX*'B, these arcs having the same sign. Br r B= X Fig. 22. If A', B', X' be the antipodes of A, B, X, the ratio of section (AB, X) is positive if X be on BA' or AB', negative if X be on AB or 'B', null,, at A or A', infinite,, at B or B'. F2 68 Various Applications. We may remark that (AB, X) = (AB, X') = (A'B', X)=- ('B, X); since it follows that when an arc AB meets another XYX', it is indifferent whether we take for the ratio (AB, X) or (AB, X'). DEF. XIX.-If A, B, X, Y be four points on the same great circle, the ratio of the two ratios of section (AB, X) (AB, Y) or sinXA sin YA. sn XA..: -_n is called the anharmonic ratio of the four points. sin XB sin YB If the ratio = - 1, the points X, Y divide AB harmonically. For example, the two bisectors of an angle of a triangle divide the opposite side harmonically. With four points on an arc of a great circlc, the same as with four points on a right line, we can, as in Sequel to Euclid, p. 127, form six anharmonic ratios, any one of which may be called the anharmonic ratio of the points. DEFINITION XX.- When three arcs of great circles a, /3, y pass through the same point 1M, and are intersected in A, C, B by thegreat M Fig. 23. circle described, with M as pole, the ratio of section (af/, y) sin CA sin (ya) sin (B szin (yP)' Theory of Transversals. 69 DEFINITION XXI.-The anharmonie ratio of four great circles (a, /3, y, 8) passing through the same point M is (a/ys3) sin (ya) sin (Sa) sin CA. sin DA sin(yyp) sin(8, )' siin CB sin:DB' If (ap3y8) = - 1, tie pencil a, /3, y, 8 is said to be harmonic. 70. If a great circle intersects the sides of a triangle ABC in the points A', B', C', then, 1~. (AB, C')(BC, A')(CA, B') = 1. (289) 2~. (ab, CC') (bc, AA')(ca, BB') = 1. (290) B '.~ ~. Fig. 24. To prove 1~-If the sides AB, CA be eut internally, and BC externally, and perpendiculars p', p", p"' be drawn to the transversal, we have by the properties of right-angled triangles, sin p' sin p" sin p (AB,C)- si-', (B C, A')=, and ( CA, B') -sin p sin " ps sin p' Hence, by multiplication, we have (AB, C') (BC. ') ( CA. B') = 1. To prove 2~-We have, by equation (55), the equalities sinA4 sin B (AB C') = (ab, CC') sinB (BC, ')= (be, AA') sin C sin C' (CB') t (ca,BB') sin. Hence, by multiplication, the proposition is proved. 70 Various Applications. Reciprocally.-If the points A', B', C' satisfy either of these relations, they lie on a great circle. Cor. 1.-The great circle which bisects two sides of a triangle meets the third side at the distance of 90~ from its middle point. Cor. 2.-The feet of two internal bisectors, and the foot of the third external bisector, of the angles of a triangle lie on the same great circle. 71. If the arcs which join the vertices of a triangle ABC to the same point O of the sphere meet the opposite sides in the points A', B', C', then, 1~. (ab, CC') (bc, AA') (ca, BB') = - 1. (291) 2~. (AB, C') (B C, A') ( CA, B') = - 1. (292) 1~. This follows from applying the theorem, ~ 29, to the three triangles A OB, BOC, COA, and considering that if the point O be inside the triangle ABC, the three ratios of section (ab, CC), &c., are negative; and if 0 be outside, two are positive and one negative. 2~ follows from 1~ by equation (55). Reciprocally, if the points A', B', C' satisfy either of the equalities (291), (292), the arcs AA', BB', CC' are concurrent. Cor. 1.-The three medians AA', BB', CC', are concurrent. Cor. 2.-The three altitudes, ha, hb, h, of a spherical triangle are concurrent. Cor. 3.-The homologous sides of two supplemental triangles intersect in points situated on the same great circle, having as pole the common orthocentre of the two triangles. Cor. 4.-The arcs which join the vertices of a spherical triangle to the points of contact of opposite sides with inscribed circle, or with any of the escribed, meet in the same point. (The GERiGONNE point of the Triangle.) Theory of Transversals. 71 72. The anharmonic ratio of a pencil (a/3y8) of four great circles (see fig., Def. ii.) is equal to the anharmonic ratio (AB CD) of the four points in which it is intersected by a transversal. DEM.-We have, equation (55), sin A4 (a3, y) = (AB, C). sin B sin A and (a3, 8) = (AB, C).. sin Hence, by division, (apt3y) = (ABCD). (293) 73. E]ach diagonal of a complete spherical quadrilateral is divided harmonically by the two remaining diagonals. Fig. 25. DEM.-Let the quadrilateral be BCB'C'; AA', BB', CC' its three diagonals. Let BB', CC' intersect in M. Join AM, and produce to cut BC in A". Now we have (Art. 70), (AB, C') (BC, A") (CA, B') = - 1, and (Art. 71), (AB, C') (BC, A') (C, B') =+ 1. Hence (BC, A") = - (BC, A');... ', A", B, C are harmonic points. Therefore (A - A'"B C) is a harmonic pencil. Hence the proposition is proved. 72 Various Applications. EXERCISES.-XXI. 1-2. If the medians AA', BB', CC' of the triangle ABC intersect in G, prove1~. sin GA - sin GA' = 2 cos 2 a. (294) sin AA' sin BB' sin CC' 2. sin A'G sin B'G sin C'G (295) 3. If through a fixed point P we draw any two transversals PAB, PA'B' to a fixed angle XSY, the locus of the points of intersection of the arcs AB', A'B is a great circle called the polar of P, with respect to the angle XSY. 4. If two spherical triangles ABC, A'B'C' are such that the arcs AA', BB', CC' are concurrent, the pairs of corresponding sides AB, A'B'; BC, B'C'; CA, C'A' intersect on the same great circle. This may be proved by transversals (see Sequel to Euclid, p. 131), or by considering the tetrahedrons (O - ABC)(O - A'B'C') cut by the same plane, which gives two rectilineal triangles in perspective. 5. If we take on the three sides of a triangle from their middle points arcs equal to a quadrant, the six points thus obtained are on the same great circle. 6. If the arcs AP, BP, CP meet the sides BC, CA, AB in A', B', C'; and if A", B", C" be the symmetriques* of A', B', C', with respect to the middle points of the sides, then AA", BB", CC" meet in the same point P', called the isotomic conjugate of P, with respect to the triangle. 7. Prove that the three arcs AD, BE, CF, each bisecting the area of a spherical triangle ABC, are concurrent.-(STEINER.) From the given conditions the spherical excess of each of the triangles BAD, CAD is E. Hence sin AD - |si s in (B-E) Isin E. sin (C- E) 2 s sin BAD. sin ADB ~ sin CAD. sin ADC. sin BAD: sin CAD::sin (B- - E): sin (C - ), from which and two similar proportions the proposition follows. * For shortness, we say that the extremities of an arc of a great circle are symmetriques, with respect to the middle of that arc. Theory of Transversals. 73 DEF. XXII. —We shall call the normal co-ordinates of a point M, with respect to a triangle ABC, quantities proportional to the sines of arcs drawn from M perpendicular to the sides of the triangle, and denote them by &a, 8b, 8. DEF. XXIII.-We shall call the triangular co-ordinates of M half the products of the sines of the perpendiculars from M on the sides, multiplied by the sines of the sides. The triangular coordinates of M are equal to the Staudtians of the triangles A MB, B CU, CXA; we shall denote them by n,, no, n0. 8. If arcs AM, BM, CM meet the sides of ABC in A', B', C', respectively, prove that (BC, A') = n,: nb; (CA, B') = na: n,; (AB, C') = nb: n,. 9. If two points be isotomic conjugates, they have reciprocal triangular co-ordinates. 10. If three arcs drawn through A, B, C be the symmetriques of any three arcs AM, BM, CM, with respect to the bisectors of the angles A, B, C, they meet in a common point M', called the isogonal conjugate of M. 11. If two points be isogonal conjugates with respect to a triangle, their normal co-ordinates are reciprocals. 12. If a transversal T cuts the sides of AEC in A', B', C', the symmetriques of A', B', C', with respect to the middle points of BC, CA, AB,are upon the same arc of a great circle T, called the isotomic transversal of T. 13. In the same case, the symmetriques of the arcs AA', BB', CC', with respect to the bisectors of the angles A, B, C, meet the sides of ABC in points which lie on the same great circle T", called the isogonal transversal of T. 14. Prove that the triangular co-ordinates of G, the point of intersection of the medians of a triangle, are equal to one another. 15. If Al, Bi, Ci be the harmonic conjugates of the points A', B', C', in which a transversal T cuts the sides of ABC with respect to the sides, then the arcs AA,, BB1, CC, co-intersect in a point r, called the trilinear pole of T. T is called the trilinear polar of r. 16. If G be the intersection of the medians, M any point of the sphere, n' the Staudtian of the triangle BGC; then cos MA4 + cos MB + cos MC A cos B c = constant = n - n'. (296) cos MG 74 Various Applications. DEM.- cos MB + cos MC = 2 cos - cos M2'. cos MA'. sin AG + cos MA. sin GA' = cos MG. sin AA'; cos MB+-cos =2cos-ca M' c os MG. sin GA 2 ( sin AG sin AG M A G B C Fig. 26. cos MA + cos MB + cos MC a sin AA' 2 cos -. cos MG= 2 c sin AG' a sin AG since 2 cos - 2 sin GA' cos MA + cos MB + cos MC sin AA' n Hence = - = - cos MilG sin GA' n' 17. Calculate the norm of the sides of BGC. We have n' sin GA' a sin (AA' - GA') a n =sin.AA, 2 cos 2 sin GA' cos b + cos c = 2 cos. cos AA. n sinAA" 2 sin GA' 2 We shall eliminate GA' and AA' between these equations. The second gives 2 cos - sin GA' = sin AA' cos GA - cos AA' sin GA'; sin GA'2 cos 2 + cos A') = cos GA. sin AA'; * - 2 cos a + cos AA' = cos GA' = 1 - sin2AA', n'2 1 1 4 cos2 + 4 cos - cos AA' + 1 4 cos2 + 2 (cos b + cos )+ 1 2 2 2 Incircles. 75 n' 1 Hence (297) Hence n ^/ 1 +_ 2(1 + cos a + cos b + cos c) 18. If tangents be drawn at A, B, C to the circumcircle of the triangle ABC, forming a triangle A'B'C', the arcs AA', BB', CC', are concurrent. The point of concurrence, K, is called the LEMOINE point of the Triangle. 19. Prove that the normal co-ordinates of K are sin (A - ), sin (B -E), sin (C - E). 20. The triangular co-ordinates of the orthocentre are tan A, tan B, tan C, and the normal co-ordinates, sec A, sec B, sec C. DEF. XXIV.-The isogonal conjugate of 0, the intersection of the medians, is called the SYMMEDIAN point. 21. Prove that the Symmedian point of a spherical triangle does not coincide with its Lemoine point. Its normal co-ordinates are sin a, sin b, sin c. 22. The normal co-ordinates of the pole of the circumcircle are cos (A-E), cos (B-E), cos(C-E). 23. If M be any point of the sphere, and the arcs MA, MB, MC meet the sides BC, CA, AB in A', B', C', if O be the pole of the circumcircle, sin MA4' sin'MB' sin MC' cos MO sin A' sin BB' sin CC' cos (STN.) (298) 24. If two equianharmonic pencils have a common ray, the intersection of three corresponding pairs of rays lie on a great circle. Compare Sequel to Euclid, Prop. v., p. 131. SECTION II.-INCIRCLES. 74. To find the radius of the incircle of a spherical triangle ABC. A F E D C Fig. 27. Sol.-Bisect the angles A, B by the arcs A 0, B 0. O is the 76 Various Applications. incentre required; and the perpendiculars OD, OE, OF on the sides are the angular radii. DEM.-It is easy to see that OD, OE, OF are all equal; also that AF = s - a. Now from the right-angled triangle OAF we have, equation (110), tan OAF= tan OF - sin AF; or, denoting the radius by r, tan A 2 = tan r sin (s - a). (299) Hence, substituting for tan - A its value, equation (22), we get tanr= sin(- ) sin(s - b) sin( - c) n (300) sin s sin s Cor. 1.-If in the expression for tan r we substitute for b, c (7r - b), (7r - c), we get the expression for the in-radius of the colunar triangle B CA' formed by producing the sides AB, A C. Hence, denoting it by ra, we get tan r, = -n (301) sin (s - a)' Similarly, tan rb =.n (302) sin (s - b)' and tan r, = in ( - (303) sin (s - c)' Cor. 2.-From the equations (299), (300) we get the following formulae for solving a spherical triangle when the three sides are given: ltan r= sin (s -a)+ Isin(s - b)+ Isin(s - c) - lsin s}. (304) ltan - l = tan r - l sin (s - a), &c. (305) DEF. XXV.-The incircles of the colunar triangles are called escribed circles. Incircles. 77 ExERCISES.-XXII. 1. Prove tan r = sin a sin ~ B sin ~ C sec A, (306) 2.,, tan r = N 2 cos 1A cos B cos C. (307) 3.,, cot r= 2N {sin(A- E) +sin (B-E) +sin (C-BE)-sinE}. (308) 4.,, if a + b + c = r, prove tan r = tan 2 A tan B tan ~ C. (309) 5.,, tan r tana ra. tan rb tan rc = n2. (310) 6.,, tan ra = sin a cos 1 B cos ~ C sec A. (311) 7.,, cotra = 2- {sin E+ sin(B - E) + sin(C- E) -sin (A - E)}. (312) 8.,, cot r - cot ra = {sin (A - E) - sin E}. (313) 9. Prove that the centre of the incircle is the orthocentre of the triangle formed by the excentres. 10. Prove cot ra + cot rb + cot rc = (cot ~ A + cot 2 B + cot ~ C) - sin s. (314) 11. Prove that the common tangents of the escribed circles taken in pairs are a + b, b + c, c + a, respectively. 12. If Oa, Oh, Oc be the centres of the escribed circles, prove that cos ra COS rb cos re cos 00a: cos 00b: cos 00c: cos ra::. (315) cos(s-a) cos(s-b) cos(s-c) 13. Prove that cot r 4 cot ra + cot rb + cot rc = { sinE+sin(A-E)+sin(B-E) +sin(C-.E)}. (316) 14. Prove sin(s-a) sin(s-b) sin(s-c) sin2. 0: sin2 BO: sin2 CO:: in sic (317) sin a sin b sin c 15. Prove that the cosines of the angles of the triangle 0,, Ob, Oc are respectively equal to cos s. sin 2 A, cos s. sin -B, cos s. sin C. 78 Various Applications. SECTION III.-CIRCUMCIRCLES. 75. To find the circumradius of a spherical triangle ABC. Sol.-Bisect the arcs BC, CA at D, E, and let O be the intersection of perpendiculars to BC, CA, at D, E; then O is the circumcentre. A Fig. 28. DEM.-Join OA, OB, OC; then, equation (113), cos OB = cos BD. cos OD, and cos OC = cos DC. cos OD. Hence OB = OC. Similarly, OC = OA; therefore O is the circumcentre of the triangle ABC. Again, the angle OAB = OBA, OBC= OCB, and OCA = OA C;.-. OCB+A -=(A + C) =S;. OCB = S - A = 90~ - - E). Let OC = R; then, from the triangle O CD we have, equation (109), cos OCD = tan DC tan OC = tan a tan R;. tanR = tan a sin(A - E); (318) and, substituting for tan a its value from equation (86), we get cot R= sin ( - E) in(B- ) sin(C- _) N sin E sin E (319) Circumcircles. 79 Cor. 1.-If in equation (319) we substitute for B, C, 7r - B, r - C, respectively, we get the circumradius of the colunar triangle BCA'. Hence, denoting it by RA, we have cot R =) (320) sin (A - E)' (320) Similarly, cot RB = (B E) (321) and cot c= sinC (322) Cor. 2.-From the equations (318), (319), we get the following formulae for the solution of spherical triangles, when the angles are given. ThusIcot R = sin(A - E) + sin (B -E) + lsin ( C - E ) - sinE). (323) i cot - a = 1 cot R - i sin (A - E), &c. (324) EXERCISES.-XXIII. sin ~ a 1. Prove tanR =. 2 a. (325) sin A cos 2 b cos 2 c 2 sin 1 a sin l b sin 1 c 2., tan R =. 2 (326) 3.t =- ^^-~ sin a(327) 3.,, tan RA (327) sin A. sin ~b. sin(2c 2 sin 1 a cos ~ b cos ~ c 4.,, tan Rz =- (328) n 5.,, tan R - tan R, = {sin s - sin (s - a)}. (329) 6.,, tan R. tan RA.tau R.tan c = N,. (330) 7.,, tan R + tan RA = cot rb + cot rc. (331) 8.,, tan RB + tan Rc = cot r + cot ra. (332) 9.,, tan R+ cot r = {tan + tan Rl + tan RB +tan Rc}. (333) 80 Various Applications. 10. Prove tanR. tan R + tan RB. tan Rc= cot r. cot ra + cotrb. cot r. (334) 11. " (cotr+ tan R)2 + 1= sina + sin b + sin 'c )2 (335) 12.,, (cotr + tan R)2 +1i = (sin b s -sin a)2 (336) 13.,, tan2]R + tan2R. + tan2RB + tan2Rc = cot2r + cot2r, + cot2 rb + cot2r. (337) 14. Prove that the angles of intersection of the circumcircle of a spherical triangle with the circumcircles of the colunar triangles are equal to the angles of the triangle. 15. The angles of intersection of the sides of a spherical triangle with its circumcircle are (A - E), (B - E), (C- E), respectively. 16. The angles of intersection of the circumcircles of the colunar triangles, in pairs, are equal to A + B, B + C, C + A, respectively. 17. If ABC be a triangle, right-angled at C, if the point C and the circumcircle of the triangle ABC be given in position, prove that the locus of the circumcentre of its colunar triangle ABC' is a great circle. 18. If b be the spherical distance between the poles of the incircle and circumcircle of a spherical triangle, prove that cos2 8 - cos2R = cos2 (R - r) - cos2 R cos2 r. (338) 19. If 5a denote the distance between the circumcentre and the incentre,of BCA', prove cos2 a4 - cos2R = cos2 (R + r,) - cos2R cos2ra. (339) 20. Prove cos = sin r sin ( sina + sin b + since ) (340) 4sinma.sin~b.s in c) ' 21. Prove cot ra + cot rb + cot rc - cot r = 2 tan R. (341) 22. In an equilateral spherical triangle, tan R = 2 tan r. 2 sin B b. sin 1 c 23. Prove that tan R sin ha = 2 sn sin (342) cos L a 24. Prove that the Lhuilierian of a spherical triangle is equal to the Lhuilierian of each of its colunar triangles, or to that of its polar triangle, <or any of the colunar triangles of the polar triangle. 25. If a, B, y, 8 denote the perpendiculars from any point in a small circle on the sides of an inscribed quadrilateral, whose lengths are a, b, c, d; prove that sin a sin y cos 2 a cos ~ c = sin j sin 8 cos b cosos 2 d. (343) 2 2 2"~ I C8I 08~d Spherical 2fean Centres. 81 26. In a quadrantal triangle, of which the side c is the quadrant, prove that cotR=sin(C-E), cotRA = sin(B-E), cot RB=sin(A-E), cot Rc = sin E. (344) 27. If the angle A of a triangle remains constant, and also the perimeter, the envelope of the side BC is a small circle. 28. If the angle A remains constant, and also b + c - a, the envelope of BC is a small circle. 29-31. Construct and resolve a spherical triangle, being given 1~. A, a, b+ c; 2~. A, a, r; 3~. A, a, R. tan (A - P-) tan (B - E) tan (C- BE) 32. Prove cos2=tan( ) +tan (B - ) +tan C-B) (345 33. Find the simplest formulae for r, ra, rb, rc; -R, RA, RB, Rc, for a diametral triangle; that is, for a triangle for which c a b C = A + B, or sin2 - = sin2 -+ sin2. 2 2 2 34. If a spherical quadrilateral be such that it is inscribed in a small circle of radius R, and circumscribed to another of radius r, prove that if 8 be the distance between the poles, sin (R + r + 8) sin (R + r-8) sin (R —r + ) sin (R- r- a) = sin4r cos4R. (STEINER.) (346) SECTION IV.-SPHERICAL MEAN CENTRES. DEF. XXVI.-If the triangular co-ordinates of a point M with respect to a triangle ABC be na, nb, n,, we have seen (Ex. xxI., 8) that the arcs AMK, BM, CM divide BC, CA, AB in the spherical ratios nc: n5, na: n,, nb: na. M is called the spherical mean centre of the points with respect to the system of multiples na nb, n,. 76. If M be the mean centre of the points A, B, C for the multiples n,, nb, n,, and P be any other point, then ncos AP + nbcos BP + nccos CP = n. cos MP. (347) DEM.-We have by Stewart's Theorem (Euc. III. 17), from the triangle BPC, cos PB sin A'C + cos PCsin BA' = cos PA' sin a; G 82 Various Applications. and from A'PA, cos PA' sin iMf+ cos PA sin MA' = cos PM sin AA'. Eliminating PA', we get cos PA sin a sin iTA' cos PB sin A'C + cos PC sin BlA' + A sin AM cos -Plsin AA' sin a sin Aix but 2no = sin AlMsin BA' sin a; 2nb = sin AJllsin CA' sin a; 2n, = sin a sin MA' sin a; 2n = sin a sin AA' sin a. B o B Fig. 29. Hence, eliminating B4', CA', MXA', AA', we get nC cos PA + nbcos PB + nPC =n cos PM. Cor. 1.-If P be a point, such that for given multiples 1, m, n, I cos AP + m cos BP + n cos CP is constant, the locus of P is a small circle. Cor. 2.-If there be any number of fixed points A, B, C, &c., and a corresponding system of multiples 1, m, n, &c., and P any point satisfying the condition 2 (1 cos AP) = constant, the locus of P is a circle. Cor. 3.-If p be the spherical radius of the circle, touching the inscribed and escribed circles of a spherical triangle (HAiT's circle), tan p = tan R. (348) DEM.-Let P be the pole of Hart's circle, I, Ia, Ib, I, the poles of the in- and circumcircles; thus IH= p - r, IaH= p + ra Ir I= p + rb, Ic= p + r,. Spherical Mean Centres. 83 Then it is evident that the staudtians of the triangles Ia IbIc Ilb Ic III,. b are proportional to 1 1 1 1 sin r' sin r,' sin rb sin r, Hence, from equation (347), we have cos (p + ra) cos(p + rb) cos (p + re) cos (p - r) _ - + sin ra sin rb sin r sin r.. 4 tan p = cot + cot rb + cot r - cot r = 2 tan R. Hence tan p = ~ tan R. lb A ITo c Fig. 30. Cor. 4.-If p7, Pb, pc be the distances from A, B, C to a great circle T passing through M, then na sin pa + nb sin Pb + nc sin P = 0. (349) This follows from (347) by supposing P to be the pole of T. Cor. 5.-If T be any great circle, and if na sin pa + nb sin p, + nc sin pc = constant, the envelope of T is a small circle. G 2 84 Various Applications. Cor. 6.-If A, B, C, &c., be any system of fixed points, and 1, m, n, &c., a system of multiples, and if Pa, Pb, Pc, &c., be the perpendiculars from A, B, C, &c., on any great circle T; then, if ( (1 sin pa) = constant, the envelope of T is a small circle. ExERCIsEs. —XXIV. 1. If from the points Ia, lb, I, perpendiculars be drawn to the sides BC, CA, AB, respectively, these perpendiculars are concurrent. 2. If H be the pole of Hart's Circle, prove that cos AH - cos = cos b cos c - cos a. (350) 3. The arcs of connection of the vertices of a triangle to the points of contact of the opposite sides, with the circles inscribed in the corresponding colunar triangle, meet in the same point, called the NAGEL point of the triangle. 4. If tl, t2, t3 be the t gents from A, B, C to Hart's Circle, prove a b c cos tl cos t2 cos t3 = cos - cos cos -. 2 2 2 5. The normal co-ordinates of the pole of Hart's Circle are cos(B- C), cos(C-A), cos(A-B). 6. The normal co-ordinates of Nagel's Point are sin2 A, sin2 B, sin2 C. 7-10 Prove the following relations:1~. sin (B + C)/sin A = (cos b + cos c)/(l + cos a). 2~. sin (B - C)/sin A = (cos c - cos b)/(l - cos a). sin (B + C) cot ~ a (cos B + cos C ) cot A 30 = cos b + cos c sin (b + c) sin (B + C) tan ~ a (cos C- cos B) tan.A cos b - cos e sin (b + c) CHAPTER V. SPHERICAL EXCESS. 77. In the preceding chapters we have made frequent use of the fumtion of the angles of a triangle, called the spherical excess. In this chapter we shall enter into further detail, and give a more systematic account of its theory than could have been conveniently given in those chapters. SECTION I.-FORMULAE RELATIVE TO E. 78. Lemma. —If the triangle BCA' be colunar with ABC, it has two sides equal to r - b, r - c, and their included angle is equal to A; therefore if 2E be the spherical excess of ABC, 2A - 2E will be the spherical excess of A'BC. Hence we have the following rule of transformation:RuLE. —In any formula containing the elements b, c, A, E of a spherical triangle, we may change the sides b, c into their supplement, and E into A - E. This rule supplies easy proofs of several propositions. 79. Let ABC be a spherical triangle; bisect BC, AC in C D A Fig. 31. A', B'. Join A'B', and produce to meet AB in D, E; let fall 86 Spherical Excess. the perpendiculars AF, BG, C-H. Then evidently the three pairs of triangles AB'F, CB'ff; A'BG, A',CH; ADF, BEG are, two by two, equal in every respect. Hence it is easy to see that the angle BAF = (A + B + C); that is, = 90~ + E; therefore DAF = 90~ - E; also DF is the complement of A'B', and DA of half AB; that is, DA = 90~ -. 2' 80. Cagnoli's Theorem.-From the values of cos f a, sin a b, sin - c (~~ 32, 33), we get sin b sin - sinA sm E= --; cos a but sin A = sin b sin ' Hence sin E= - (351) 2 ccos a b c os b cos This is CAGNOLI'S Theorem. 81. By the transformation of ~ 78, we get sin (A - E) = 2 cosan (352) cos 1 a sin - b si n Hence, by interchange of letters, sin (B - E) = 2sin cosb sin c (353) 2 sinL a cos b b sin c sin (7 - E) = --— ^- --- (354) sin(C- sin - a s in b cos o ( 54 82. To find the value of cos E. From the triangle DAF, ~ 79, we get sin DAF = sin iFD sin AD, cos A'B' or cos E=; (355) Cos -c Formulae Relative to E. 87 but cos A'B' =cos a cos b + sin a sin 2 b cos C, from the triangle A'B' C. Hence cos - a cos 2 b + sin 2 a sin 1 b cos ( ~~cos.E= _____cos_ C *os co-E —. (356) COS 2 C And substituting for cos C its value from equation (15), and reducing, we get E cos2-a + cos2- b + cos21 C - 1 2 cos - a cos - b cos (3 Or thus: cos b cos AA' cos - acc~ b os cos -' + cos 2 cos 'B' cos -. 2 2 Hence, eliminating cos AA', we get a 2 cos b + cos c C,osi' 2 1 + cos a + cos b + cos c a b a b 4 cos - cos - 4 cos - cos Hence from (355), (358) 1 + cos a + cos b + cos c cos E = a b (359) 4 cos - cos cos2 2 2 Cor. 1.-From (351) and (354) we get cota cot b cot= co + cot C; (360) sin C and by interchange of letters we get cot E cot cot -. (361) sin A and cot - c cot - a cot E + cot B. (362) sin B Cor. 2.-If the area of a spherical triangle and one of its angles be given, the product of the semitangents of the containing sides is given. 88 Spherical Excess. 83. By the transformation of ~ 78 we get, from (357), cos2 1 a + sin2 2 b + sin2 c - 1 cos (A - E) =: a b c 2 cos - sin sin 2 2 2 ior cos - ) =i 1b + sin - - sin2 a or cos({A - E)= 2-b — o (363) 2 cos - a sin b sin Similarly, sin2a + sin2 Sa, - sin21 b cos (B - E)= sinas sin (364) - sin a cos Il b sin ' and cos(C-)= sin2 - a + sin2 b - sin2 C( co2 s in a sin b cos (365) 2sinlasin-Lbcosl From the formulae (360)-(362) we get, by transformation, the following values for cot (A - E), viz., cot atan bb cot(A - E) cot ta cot C (366) sin C tan a b tan c ta2-b tan + cot A (367) sin - tan,a cot } b tan cot- - cot B, (368) sin B with similar values for cot (B - E), cot ( C - E). 84. To find sin OE, cos E, tan -E. sin = I -cos E 2 E= 2 i-^ ~ 4 cos 2 a cos 1- b cos - c (from (357)) -sin 1 s sin 1- (S - a) sin - (s - b) sin 1 (s - c) (369) cos 1 a cos 2 b cos- 2 Similarly, /coS -S C cos 2 (s - a) cos ( - ( b) cos (s - c) (370) cos ojG I -1 i 3 — 7 --- ---- - - i 2,, î acos - a cos ob cos - c Hence tan LE= /tan t s -tan(- tan 1 (s - b tan 1 (s-c). (371) Formulae Relative to E. 89 The same value as that obtained in ~ 48 by a different method. 85. If we put L = V/cot 8 s tan - (s - a) tan (s - b) tan (s - e) (see equation (186)), we have L tan E = t (372) cot -îs Hence (~ 78), L ta - ) = tan ( ' (373) tan (B - ) = tan (- b) (374) tan (C - E) = La (375) Compare equations (182)-(185.) 86. Lhuilier's theorem can be proved, independently of sin — E, cos-1 E as follows. Thus:sin (4 + + C-7r) sin (A4 + B) - sin (w - C) c2 os (A +B + C- r) cos- (- + B) +cos (7r - C) sin 1 ( + B) -cos C cos (a - b)- cos c cos C- C cos 2 (A + B) + sin 1 C~ cos'i (a + b) + cos 1 c * sin -1 C (by Delambre's Analogies) _sin 1 (s - a) sin 1 (s - b) cot C cos 3- s cos ( (s- c) = /tan 2 s tan 1 (s - a) tan - (s - b) tan 2 (8 - c). 87. Prouhet's proof of Lhuilier's theorem. From the third of Delambre's Analogies we have sin co si 2 — -c Sm- cos2 2 90 Spherical Excess. therefore C. C a + b sm 2 + sinm 2 - E cos + cos * c^ /C7 ~\ c a+b sin ' - sin E cos 2 - cos 2 2 22 y 2 Hence tan - E cot ( C - E) = tan tan - (s- c). (376) Similarly, from the first of Delambre's Analogies, tan i lEtan î (C-E) = tan - - (.Etan t -(s-b). (377) Ience, multiplying and extracting square root, &c. See Nouvelles Annales, 1856, p. 91. EXERCISES XXV. 1. If the arc AD, drawn from A to a point D in the side BC, bisect the area of the spherical triangle, prove that cos AB: cos AC: sin ~ B1): sin 2 DC. (378) 7r 1r i 2. If ABC be a triangle, having a = b =, c =, prove sin = B. 3 2 3 3. If in fig., ~ 79, the arc EM be cut off equal to AB4, and MN be drawn perpendicular to the great circle DE; then MN= E. (GUDERMANN.) cos 1 b cos 1 c sin A 4. Prove that sin (A - E) = cs 2 b c c sin (79 cos2a 5. Prove that cs ( - E in \ b sin ~ c + cos 1 b cos 1 c cos cos(A - cs) = (380) cos a 6. Prove that in a right-angled spherical triangle tan E = tan ~ a tan ~ b. 7. If a', b', c', ', B', C' denote the sides and angles of the triangle supplemental to ABC, prove cot e s cot s' = tan (s - a) tan ~ (s'-a'). (381) 8. In the same case, if 2E' denote the spherical excess of the polar triangle, prove that tan LE tan ~ E' = tan (A - E) tan (A' - E'). (382) 9. Prove that the arc joining the middle points of any two sides of a spherical triangle is less than a quadrant. Formulae Relative to E. 91 10. The cosines of the arcs joining the middle points of the sides of a spherical triangle are proportional to the cosines of half the sides. 11. Solve a spherical triangle, being given a, b ~ e, and E. 12. If s denote the semiperimeter of a spherical triangle, A, a,, A, A its area, and the areas of its colunar triangles; prove that tan2 - = tan a A. cot A. cot b A. cot - ac. (383) 13. Prove sin E = cot R tan a tan b tan 2 c. (384) 14.,, sin s = sin a cos B cos C sin A. (385) 15. If (a + b + c) = 7r, prove cos A + cos B + cos C 1, cos A. (385) B C 16. In the same case, prove that cos A = tan - tan -. (386) 2 2 sin21A + sin2 B +- sin2 C!- 1 17. Prove coss=.+ (387) 2sin A sin i B sin ( C 8., sin sinEcos(A-E) cos o(B-E)cos2C-/) (388) 2 4 sin ~A sin B sin ~ C b c 19. If E = r, prove that cosA = - cot - cot -. (389) = 2z%' 2 o2 cos2B+ cos21 C - cos2. 20. Prove cos (s - a) = 2 cos 2-c 2 (390) 2 sin A cos B cos ( C 21. If I be incentre of a spherical triangle, prove that A B C Cos2 - os2 - - cos2 - 2 2 2 cos BIC = B C. (NEUBEaG.) (391) 1B C 2 cos - cos - 2 2 22. If Ia, Ib, Ic be the incentres of the triangles colunar to ABC, prove that A4 B C cos2- - sin2 -sin2 cos BIaC = - - (Ibid.) (392) 2 sin - sin - 2 2 23. The angle BI,, C corresponds in the supplemental triangle to the are joining the middle points of two sides. (Ibid.) 24. If Ia be the incentre of the colunar triangle A'BC, from Ia let fall perpendiculars IaD, IE, IF on the sides BC, CA, AB, respectively; then the angle BIa C = ~ FIE = FIaA. The triangle FIaA gives cos FI A = cos AF. sin FAIa = cos s. sin. A. Hence cosBI C = co s. sin ~ A. 92 Spherical Excess. Hence, from (392), we get A 2B cos2 — smn - sin2 -COS s = 2 si n 2 si n 2 (NEUBERG.) (393) 2 sin 1 A sin - - sin ' C' This theorem is the correlative of 357. We can get, in the same manner, cos (s-a), cos(s-b), cos(s-c); cos, sin tan tan2 &c. SECTION II. —LEXELL'S THEOREEM. 88. If the base BC of a spherical triangle ABC be given in magnitude and position, and the spherical excess in magnitude, the locus of the vertex is a small circle of the sphere. STEINER'S PROOF. Lemma.-If upon the base BC a spherical triangle be constructed, such that A - E is given, the locus of A is a small circle, namely, the circumcircle of the triangle. For if O be the pole of the circumcircle (see fig., ~ 75), the angle OBC= OCB = (A - E). Hence O is a given point, and the circle is given in position. I,; LEXELL'S THEOREM.-Let AB C be one position of the triangle, 2E the spherical excess constant. Let the points B', C' be the P, Fig. 32. antipodes: of B, C; let P be the pole of the circle AB'C'; then we have 2 E=A + B + C- 7r = B'AC'+ r - AB'C' + wr Lexell's Theorem. 93 - A C'B' - r = r A - B'- C'. Hence B' + C'- A is constant; and by the lemma the locus of A is the circumcircle of the triangle B'C'A. Or thus: SERRET'S PRooF.-Let, as before, B', C' be the antipodes of B, C; let E' be the spherical excess of B'C'A, and R' its circumradius; then we have (~ 75), tan R' = tan a a ' sin (A - E') = tan a a ' sin E. Hence since a and E are given in magnitude, R' is given in magnitude, and the circumcircle of B'C'A is evidently given in position, and is the locus required. 89. Steiner's Theorem.-The great circles through angular points of a spherical triangle ABC, and which bisect its area, are concurrent. Let the circles bisecting the area meet the opposite sides in the points a, /, y, respectively; also, let A', B', C' be the antipodes of A, B, C. Now the areas of the triangles ABa, AB,3 are equal, each being half of ABC. Hence, by Lexell's theorem, the points A', B', a, / are concyclic. Similarly, each of the systems of points B', C', /, y; C', A', y, a, is concyclic. A B C Fig. 33. Let the point common to the planes of these three small circles be P, then the lines of intersection of these planes two by two pass through P. Hence, if O be the centre of the sphere, the planes OB'f3B, OC'7yC, OA'aA have a common line of intersection, namely, the line OP. Hence the proposition is proved. 94 Spherical Excess. Analytical Proof.-The triangles ABa, A Ca having the same spherical excess, we have, by Cagnoli's theorem, ~ 80,. c.Ba. b. Ca. sin si - sin B sin sin - sm C 2 2 2 2 cos Aa cos i Aa Hence Ba c sin- cos - 2 2 Ca- b sin cos and from this and two similar equations we get sin 2Ba sin - CP n A7 = sin A = a sin RaC sin 2 -B. (1) Also the triangles ABa, 7yBC having equal areas, tan c tan i Ba = tan yB tan a. (Art. 81, Cor. 2.) Hence, tan 1 Ba tan I C,/3an n -yA = tan 2 aC tan i/3A tan -yB. (2) From (1) and (2) we have cos2 Ba cos CP cos Ay= cos -aC cos BA cos -yB. (3) From (1) and (3) we get sin Ba sin Cf3 sin Ay = sin a C sin,3A sin yB. (4) Hence the arcs Aa, Bl, Cy are concurrent. (NEUBERG.) Cor.-The triangular co-ordinates of the point of intersection of the arcs Aa, B/3, Cy are cos a - cos b cos c, cos b - cos c cos a, cos àC - cos a cos b. (Ibid.) These values are obtained from the equation sin Ba cos - c sin ( a - j-Ba) cos b' which gives cot 2Ba, and thus sinBa. Lexell's Theoren. 95 90. Keogh's Theorem.-The sine of half the spherical excess is equal to twice the Staudtian of the triangle formed by joining the middle points of the sides.-Nouv. Annales, 1857, p. 142. DEM.-Let A',',, C' be the middle points of the sides. (See figure, ~ 79.) Then we have, from the right-angled triangle DAF, cos DAF = sin D cos DF, equation (111); that is, sin E = sin D sin B'A'. But sin D = sine of the perpendicular from C' on B'A'. Hence, if n' denote the Staudtian of A'B'C', we have sin E = 2n'. (364) 91. To find the triangle of maximum area, two sides, b, c, being given. /A C c Fig. 34. Sol.-The following is Steiner's geometrical solution: Suppose the side AC to be fixed in position. Let A'C' be the antipodes of A and C; through A', C' let a small circle be described with pole P, such that the angle PA'C'= - -; then every triangle having A C as base, and vertex any point on 96 Spherical Excess. the small circle, will have a constant area, namely, 2Er2. If the side AB be given, the small circle BB1 described with A as pole, and spherical radius c, cuts Lexell's circle in the points B,,, each of the two triangles AB C, AB, C will have an area = 2Er2. In order that the problem may be possible, Lexell's circle must meet the circle BB1; or, what is the same thing, the angle PA'C' equal to - - E, must be sufficiently large, the minimum of - - E, or the maximum of E, corresponding to the case where the small circles touch each other. Then the points A, B, P, A' are on the same great cirele, and the triangle BC'A' is a diametral triangle;.. C' = '+ B; but A'= r - A, and C'= 7r - C. Hence A = B + C, and the required triangle is diametral, a being the diameter. Cor.-If AB be greater than A C', the circle BB, must intersect Lexell's circle, and there will be neither a maximum nor a minimum; but if AB be greater than A C', AB + A C will be greater than AA ', or b + c greater than 7r. HIence, if b + c > w, there will be neither a maximum nor a minimum. Trigonometrical Solution (NEUBERGG'S). 1~. We have, by Cagnoli's formulae (351), (352), sin (A - E) = sin E cot b cot i c. If cot bcotlc> 1, or b+c>180, sin E may have any value, and then sin (A - E) may be found, and the triangle is possible. Hence there is neither a maximum nor a minimum. Lexell's Theorem. 97 2~. If cot b.cotlc=l, or b+c=-r, we have A - E = E, and the triangle becomes a lune formed by the circle C'AC, and the greater circle tangential to Lexell's circle in C'. 3~. If cot} b.cotc < 1, or b + c < 7, sin E is no longer arbitrary. In order that sin (4 - E) < 1, sin E must be <tan b tan-c. The maximum of E corresponds to sin E = tan b. tan c, and then A-E=, or A=B +C. 2' EXERCISES.-XXYI. 1. Construct a lune equal in area to a given triangle (make use of Lexell's circle). 2. Construct by means of Lexell's theorem a triangle ABIC' equal in area to a given triangle ABC, and having two given sides bl, cl. 3. Construct on the side BO of a given triangle ABC an equivalent isosceles triangle. 4. Convert a triangle ABC into an equivalent isosceles triangle, having a common angle A. 5. Transform a spherical polygon ABCDE into an equivalent spherical triangle. [Employ Lexell's circle in the same manner as parallel lines are employed in the corresponding question in Plane Geometry.] 6. Being given a spherical polygon ABCDE, if the sides be produced in the same sense, and with each vertex as pole, an arc of a great circle be described, limited by the sides of the corresponding exterior angle of the polygon, prove that the total figure thus formed is equal in area to a hemisphere.-(NEUBERG. ) 7. Being given A and E, prove that, if a is a minimum, b = c. sn sin b sin o sin A We have sin E = "si b sin COS 2 Hence, from the required condition, sin ~b sin c is a maximum; but H 98 Spherical Excess. tan b tan c is constant (~ 82, Cor. 2); therefore cos b cos c is a maximum;. sec b sec c = (tan b - tan c)2 + (1 + tan ~ b tan c)2 is a minimum. Hence b = C.-(NEUBERG.) 8. Being given A and E, prove that if b + c be a maximum, b = c, tan ~ (b + c) = 1tn Hence b = c.-(Ibid.) 1 - tan b b tan e c 9. If ABC, ABD be two spherical triangles of equal areas on the same side of the common base AB, prove that sin - AB. sin C~D + cos AC. cos 2 BD = cos ~ AD. cos BC. 10. Investigate the maximum or minimum of E, being given A and b + c. cot b. cot c ( 2 cos (b + c) cot.E =.1 + — T -,. + cotA sinE A i cos (b-c)-cos~ (b+c) sin+ctA 2 cos ~ (b + c) a + cot sinA {cos (b - c) - cos i (b + c) } 2 If ~ (b + c) < 90~, then cos ( - c)cos ( + ) > 0, and the minimum of cot E or the maximum of E corresponds to cos (b - c) = 1 or to b = c. If - (b + c) > 90~, in the colunar triangle A'BC, ~ (b' + c') < 90~; and since A and b + c are constant in ABC, A and b' + c' are constant in BCA'. Hence the area of BCA' is a maximum when b' = c'; and therefore when ~ (b + c) > 90~, the area of ABC is a minimum when b = c.-(Ibid.) 11. If E= -, prove that cos A = - cot ~ b cot ~ c. 12. If O be a point such that the areas of the triangles AOB, BOC, COA are equal, prove that tan 2 n: t an:: sin BOC- 3: sin( -:sin A OB — ) 13. In the same case prove that the small circle passing through the antipodes of 0, and the extremities of any side of the spherical triangle, intersects that side at an angle = E. 3. CHAPTER VI. SMALL CIRCLES ON THE SPHERE. SECTION I. —COAXAL CIRCLES. 92. If an arc of a great circle passing through a fixed point O cut a small circle X in the points A, B, tan - AO. tan - OB is constant. -\~ c \ Fig. 35. DEM.-Let P be the pole of the small circle. Join OP. Let fall the perpendicular PC; then, from the triangles A CP, O CP, we have cos A C cos AP cos CO - cos OP' Hence tan 4 (A C + CO) tan ~ (A C - CO) = tan (AP + PO) tan i(AP - PO); or, denoting the radius of X by p and OP by 8, tan i OA. tan % OB = tan 2 (p + 8) tan (p - 8). 395) 112 100 Small Circles on the Sphere. DEF. XXYII.-lThe product tan 2 OA. tan 2 OB is called the spherical power of O, with respect to the circle. It is positive or negative, according as O is exterior or interior to the circle. Cor. 1.-If 8 be the distance of a point O from the pole of a small circle, radius p, the spherical power cos p - cos 8 (396) cos p + cos 8 Cor. 2.-If from any point O outside a small circle two arcs be drawn to it, of which one, OD, is a tangent, and the other a secant, meeting it in the points A, B; then tan2 ~ OD) = tan OA. tan i OB. 93. If two small circles eut orthogonally, the plane of either passes through the vertex of the cone, touching the sphere along tle other. Fig. 36. DEM.-Let the circles be X, Y; O, 0' their spherical centres, A, B their points of intersection; then it is evident that the tangent line to the arc AO is in the plane of Y, and that it passes through the vertex of the cone, which touches the sphere along the circumference of X. Hence the proposition is proved. Cor.-If any number of circles on the sphere have a common orthogonal circle, their planes pass through a common point; Coaxal Circles. 101 and conversely, if the planes of any number of circles pass through a common point, they have a common orthogonal circle. 94. If the planes of a system of circles S have a common line L of intersection, the circles S have an infinite number of common orthogonal circles. DEM.-Take any point P in the line L, and through it draw tangent lines to the sphere; these will touch along a circle which (~ 93) cuts each circle of the system S orthogonally; and since the same thing holds for each point on L, we have an infinite number of circles forming a system S', each of which cuts each circle of 8 orthogonally. Cor.-The planes of the circles of the system S' have a common line of intersection. For, take any two of them, say P and Q. Now (~ 93) the plane of each passes through the vertex of each of the cones, touching the sphere along the circles of the system S. Hence the vertices are collinear, and the plane of each circle of S' passes through the line of collinearity. DEF. XXVIII.-A system of circles S, whose planes pass through a common line L, is called a COAXAL SYSTEM. DEF. XXIX.-The circle of the system S, whose plane passes through the centre of the sphere, is called the RADICAL CIRCLE of the system. DEF. XXX.-If through L two tangent planes be drawn to the sphere, their points of contact, regarded as infinitely small circles, are the limiting points of the system. Cor.-Each circle of the system S' passes through the limiting points of S. 95. If X, Y, Z be three circles of a coaxal system, and from any point P in X tangents PT, PT' be drawn to Y and Z; then sin PT: sin - PT' ir a given ratio. 102 Small Circles on the Sphere. DEM.-Let O, O' be the spherical centres of Y, Z. Join OP, O'P by arcs of great circles; then, if the radius of the sphere be unity, the perpendicular from P on the plane of Y= cos OT - cos OP = cos OT - cos OT. cos PT = cos OT. 2 sin2 PT. Similarly, the perpendicular from P on the plane of Z= cos O'T. 2 sinU PT'. But since the planes of X, Y, Z are collinear, the perpendiculars have a given ratio. HIence the ratio of cos OT. sinm PT: cos O'T'. sin2 PT' is given, and OT, O'T' are given, being the spherical radii of Y and Z. Hence the ratio of sin PT: sin J-PT' is given. Cor.-If PT = PT', the locus of P is the radical circle of the system. EXERCISES.-XXVII. 1. The radical circles of three small circles taken in pairs are concurrent. 2. If there be a coaxal system of circles S, and a circle X distinct from it, then the radical circles of X, combined with each circle of S, are concurrent. 3. If through a point on the radical circle of two small circles we draw a spherical secant to each, the four points of intersection are concyclic. 4. Through two points of the sphere describe a small circle touching a given great circle. 5. If through a fixed point A we draw a great circle, cutting a given small circle in the points B, C, and if a point D be taken on it, such that tan2 ~ AD = tan2 DB. tan2 1DC, prove that the locus of D is a great circle. 6. If X, Y be two small circles; PT, PT' two tangents to them from a point P, prove that the locus of P is a circle, if m cos PT + n cos PT' be constant, m and n being given numbers. 7. The locus of the poles of small circles, intersecting two small circles X, Y at the extremities of two spherical diameters, is the radical circle of X, Y. 8. Describe a circle cutting three small circles at the extremities of three spherical diameters. Centres of Similitude. 103 9. If two rectangular secants intersecting in M cut a small circle in the pairs of points A, B; C, D, prove that tan2 M + tan2 MB + tan2 MC + tan2XD) = M OSp + COS (397) (cos p + cos 3)2 where 8 is the distance of M from the pole of the small circle.-(NE3uBERG.) 10. If from any point P of a great circle MP tangents PT, PT' be drawn to a small circle, prove that tan MPPT. tan ~ MPT' is constant. 11. The difference of the cosines of the tangent arcs, drawn from any point P on the surface of a sphere to two small circles X, Y, is proportional to the sine of the perpendicular drawn from P to the radical axis of X and Y. SECTION II. —CENTRES rF SIMILITUDE. 96. DEF. XXXI. —Two points, S, S', which divide the arc PP', joining the poles of twuo small circles Y, Z externally and internally in the spherical ratio of the sines of the radii, are called the centres of similitude of the small circles. MP/ \Fig. 37. \ Fig. 37. Cor.-Common tangents to the small circles pass through the centres of similitude, viz., the direct common tangents through the external centre, and the inverse common tangent through the internal centre. DEF. XXXII.-If through a centre of similitude we draw a secant cutting the circles, then the pairs of points M, M'; N, N' are said to be homothetic, and M, N'; 1f', N are inverse. 104 Small Circles on the Sphere. 97. If the secant through a centre of similitude S meets the circles in the homothetic points XM, M; then tan ~ SM: tan SNM' in a given ratio. DEM.-From the definition sin SP: sin PM:: sin SP': sin P'f. Hence it follows that the angle SMIP = SM'P'. Now since the triangles SMP, 3SM'P' have two angles in one respectively equal to two angles in the other, it follows from the third of Napier's Analogies that tan: SM: tan 1 SM:' tan 1 (SP +P P) tan 1 (SP '+P'1T); that is, in a given ratio. Similarly, tan L S N: tan ' SN' in a given ratio. (398) Cor.tan 1 SM f. tan 1 SN' is constant, as also tan i SN. tan 1 SM'. This follows from ~~ 92, 97. (Compare Sequel, Prop. II., page 83. Cor. —If there be given a point S and a circle Y, and on the arc Sf joining S to any point MJon Y a point N' be taken, such that tan 2 S1. tan 1 S7V' is constant, the locus of N' is a circle. 98. The six centres of similitude of three small circles taken in pairs lie three by three on four great circles, called axes of similitude of the small circles. DEM.-If a, b, c be the radii of the circles; A, B, C their spherical centres, A'B'C' the internal centres of similitude, and A"B"C" the externals; then we have by definitions (AB, C") = a - b, (BC, A")= b — c, (CA, B") = c a. Hence (AB, C"). (B C, A"). ( CA, B") = 1. Hence (~ 70) the points A", B", C" lie on a great circle. Similarly, it may be shown that any two internal centres and an external centre lie on a great circle. Centres of Similitude. 105 Cor. 1-If a variable circle touch two fixed circles, the great circle passing through the points of contact passes through a fixed point, namely, a centre of similitude of the two circles; for the points of contact are centres of similitude. Cor. 2.-If a variable circle touch two fixed circles, the tangent drawn to it from the centre of similitude, through which the chord of contact passes, is constant. 99. DEF. XXXIII.-Being given afixed point S, and any line whatever, y, on the sphere, if upon the arc of a great circlejoining S to any point MW of y a point M1' be taken, such that tan SM: tan SM' in a given ratio, the locus of M' is said to be homothetic to y. DEF. XXXIV.-If JM' be taken, such that tan ~ SM1. tan SM' is constant, the locus of M' is called the inverse of y. This method of inversion was first employed in the Author's Memoir on Cyclides and Sphero-quartics. (Read before the Royal Society in 1871.) EXERCISES.-XXVIII. 1. If two small circles touch two others, the radical axis of either pair passes through a centre of similitude of the other. 2. The figure homothetic to a circle is a circle. 3. The inverse of a circle is a circle. 4-5. S being the centre of similitude of two circles; M, N two inverse points on these circles-1l, the tangents at M and N intersect on the radical axis; 2~, these points are points of contact of two circles touching the two given circles. 6. The angle of intersection of two circles on the sphere is equal to the angle of intersection of the circles inverse to them. 7. Any two circles can be inverted into two equal circles. 8. Any three circles can be inverted into three equal circles. 9. If two circles be the inverses of two others, then any circle touching three of them will also touch the fourth. 106 Small Circles on the Sphere. 10. If two points be the inverses of two other points, the four points are concyclic. 11. If through the centre of similitude S (see fig., ~ 96) another great circle be drawn, intersecting the circles Y, Z in the points g, y'; v', v, corresponding to the points M, M', N', N, the systems of points M, ', j, v'; M',.N, z', V; M, N', N ',,; M', N, l, v', are each concyclic, and the planes of the four circles pass through a common point. 12. If a variable circle on the sphere touch two fixed circles, the sine of its radius has a constant ratio to the sine of the perpendicular drawn from its spherical centre to the radical axis of the fixed circles. 13. If a variable circle touch two fixed circles, the ratio of the sines of half the tangents drawn to it from the limiting points is constant. 14. If a variable circle touch two fixed circles of a coaxal system, it cuts any circle of the system at a constant angle. 15. The inverse of a coaxal system is a coaxal system. 16. The inverse of a system of great circles passing through two common points is a coaxal system. 17. The inverse of a system of small circles having a common spherical centre is a coaxal system. SECTION III.-POLES AND POLARS. 100. LEMMAS. —If the segment AB be harmonically divided in the points C, D; and E the middle point of AB; then 1~. tan2EB = tan EC. tan FD. (399) 20. cot AB = (cot A C + cot AD). (400) ^E B Fig. 38. For, by definition, sin CA sin DA sin CB sin.DB; sin CA - sin CB sin DA - sin DB sin CA + sin CB sin DA + sin DB' Poles and Polars. 107 tan EC tan EB Silence lIIence tan EB tan ED' Hence the proposition is proved. To prove 2~-We have sin BC. sin AD = sin CA. sin DB, or sin (AB - A C) sin AD = sin A C sin (AD - AB). IIence, expanding and dividing by sin AB sin A C sin AD, we get cot A C - cot AB = cot AB - cot AD, or cot AB = - (cot A C + cot AD). DEF. XXXV.-Being given a small circle X, spherical centre P: if C, D be two points dividing the spherical diameter AB harmonically, an arc DD' of a great circle through one of these points, D, perpendicular to the diameter AB, is called the harmonic polar of the other, C, and C is called the harmonic pole of the arc DD'. D| Fig. 39. 101. The arc of contact of spherical tangents, drawn from an exterior point D to a small circle X, is the harmonic polar of D. DEM.-The right-angled triangles DEP, ECP give tan EP tan CP cs tan DP tan EP 108 Small Circles on the Sphere. Hence tan2EP or tan2EA = tan DP. tan CP; therefore the points C, D are harmonic conjugates to A and B; and therefore, &c. 102. If a spherical chord AB of a small circle Xpass through a fixed point C, the locus of the intersection of tangents AD, BD is the harmonic polar of C. F P B Fig. 40. DEM.-Let P be the spherical centre of X. Join PD, PA, PC by arcs of great circles, and let fall the perpendicular DF on PC, produced if necessary. Now because in the spherical quadrilateral DECF the angles E, F are right, we have, equation (134), tan PC. tan PF = tan PE. tan PD.= tan2P2. Hence the proposition is proved. Cor. 1.-If a variable point move along an arc of a great circle, its harmonic polar passes through a given point. Cor. 2.-If C be a fixed point in a small circle X; A C, CB any two arcs of great circles at right angles to each other; the chord AB passes through a fixed point. Poles and Polars. 109 DEM.-Let Y be the circumcircle of the colunar triangle A C'B, and 00' the spherical centres of X and Y; and since the angle C is right, the circles X, Y cut orthogonally; therefore A 0', O'B are tangents to X. Hence AB is the polar of c C Fig. 41. 0', with respect to X; and since O'A = O'C', the locus of O' is the radical axis of the circle X and the fixed point C'; and therefore AB, the polar of O' with respect to X, passes through a given point. 103. Every secant ( A0) passing through a given point O is eut harmonically by the circle X and the harmonic polar of O. Fig. 42. DEM.-Let BC be the polar of 0, and let the sines of the 110 Small Circles on the Sphere. perpendiculars from A on the sides CO, OB, BC of the triangle OB C be denoted by x, y, z; and the sines of the perpendiculars from A' by x', y', ', respectively, then we have x::: sin 0CA: sinA CB; that is, x::: sin(B - E): sin C. Similarly, y: s:: sin ( C- E): sin B; xy sin (B - E) sin (C- E) =cos.** - == -—:-p. ^ - = COS2 ' a. z2 sin B sin C xiy' In like manner, - = co2 a. z,2 Hence xy: x'y': 2: 2; but xy: x'':: sin2 OA: sin2 OA', and z2: z'2:: sin2AN: sin2A'N;. sin OA: OA':: sin AX: sin NA'. (401) EXERCISES.-XXIX. 1. If four points A, B, C, D lie on a great circle a, their anharmonic ratio is equal to that of their harmonic polars, with respect to any small circle X. For if O be the spherical centre of X, P the harmonic pole of a, the perpendiculars from P on the circles OA, OB, OC, OD will be the harmonic polars of A, B, C, D, and will pass through the poles A', B', C', D' of the great circles OA, OB, OC, OD. Now it is evident that (P- A'B'C'D') = (A'B'C'D') = (O - ABCD) = (ABCD). 2. If A, B, C, D be four points on a small circle X, and if the arcs AB, BC, C'D, DA be denoted by a, b, c, d, respectively; then if P be any variable point on X, the anharmonic ratio (P- BCD) = sin 2 a. sin c sin s b sin 2 d. For if the perpendiculars from P on AB, BC, &c., be a, B, &c., and the Poles and Polars. 111 staudtians of the triangles PAB, PBC, PCDZ, PDA being denoted by na, nb, nc, nd, we have evidently (P- BCD) = = sin a sin a. sin c sin y - sin b. sin j. sin d. sin = sin ~ a sin ~ c - sin 1 b. sin d (equation (343)). (402) 3. If A, B, C, -D be four points on a small circle X, the spherical triangle whose summits are the points of intersection of the arcs AB, CD; BC, DA, and CA, 1DB, is such that each side is the harmonie polar of the opposite vertex. This is called the harmonic triangle of the four points. 4. The harmonic polars of any point on the radical circle of two small circles with respect to these circles intersect on the radical circle. 5. If X, Y are two small circles, Z a great circle perpendicular to the great circle passing through the spherical centres of X, Y, the harmonic polars of any point of Z intersect on a great circle. 6. If a spherical quadrilateral be inscribed in a small circle (X), and at its angular points arcs of great circles be drawn touching X, their twelve points of intersection lie four by four on the sides of the harmonic triangle. 7. PASCAL's THEOREM.-If a spherical hexagon be inscribed in a circle, the opposite sides intersect in pairs on a great circle. 8. A, B, C; A', B', C', are two triads of points on two great circles; prove that the intersections of the three pairs of arcs AB', A'B; BC', B'C; CA', C'A lie on a great circle. 9. SALMON'S THEOREM.-Given any two points A and B and their harmonic polars, with respect to a small circle X, whose spherical centre is O. Let fall a perpendicular AP from A on the polar of B, and a perpendicular BQ from B on the polar of A; then, if A', B' be the harmonic conjugates of A, B, with respect to X, prove that cos OA: cos OB:: sin OA' sin AP: sin OB' sin BQ. 10. BRIANCHON'S THEOREM.-If a spherical hexagon be described about a small circle X, the three arcs joining the opposite angular points are concurrent. 112 Small Circles on the Sphere. SECTION IV.-MTJTUAL POWER or Two CIRCLES. 104. If a, /, y be the arcs connecting any point P to the vertices A, B, C of a trirectangular triangle, then cos2a + cos2 p + cos2y = 1. DEM.-Since the triangles PAB, PAC are quadrantal, we have, from (equation (154), cos, = sin a cos BAP, cos y = sin a cos CAP. A Fig. 43. Hence cos2/P + cos2y = sin2a;. cos2a + cos2/P + cos2y 1. (403) Cor.-If x, y, z be the perpendiculars from P on the sides of the triangle ABC; x, y, s are respectively equal cos a, cos /, cos y. 105. If a, P, y, a', /3', y be the angular distances (fig. 44) of two points P, P' from the vertices of the trirectangular triangle ABC; then cos PP' = cos a cos a' + cos 3 cos /3' + cos y cos y', cos PP' cos a cos a' + sin a sin a' cos PAP' = cos a cos a' + sin a sin a'(cos PA C cos P'A C+ sinPA C sin P'A C) = cos a cos a' + cos / cos,' + cos y cos y' (equation (154)). (404) Mutual Power of Two Circles. 113 A Fig. 4 Fig. 44. DEF. XXXVI.-The product of the cosines of the spherical radii of two circles, subtracted from the cosine of the arcjoining their spherical centres, gives a remainder, which is called the mutual power of the two circles.? If the circles be denoted by letters with suffixes, we shall denote their mutual power by the suffixes. Thus the mutual power of the circles Sa, sg shall be denoted by af/. 106. Frobenius's Theorem.-If s,, s5, s3, 84, s5; 8s1, 821> s3, s4, s8' be any two systems of five circles on the sphere; then 11', 12', 13', 14', 15' 21', 22', 23', 24', 25' 31', 32', 33', 34', 35' = 0. (405) 41', 42', 43', 44', 45' 51', 52', 53', 54', 55' DEM.-Let xi, y,, sz, &c., denote the normal co-ordinates of the centres of the circles, with respect to a fixed trirectangular * The introduction of this term into Geometry is due to DAIBOUX, "Annales de l'École Normale Supérieure," vol. I., 1872. I 114 Small Circles on the Sphere. triangle, and r,, r2, &c., their spherical radii; then, multiplying the determinants 0, xI, y2, Y 1, cosr' 0, X1 yt, s1, - cos r' 0, X2, Y2, Z2, COS 72 0, X2', 2 ', y- COS 2' 0, X Y3, 3, COS r3, O, X3'/. Z3/ - COS r3 0, X4, Y4, COS r4 0, x4', z4', -cos r4' 0, *X5> Y5, Zg5 COS r5 0, 51 y5/, z5', -cos r5 the proposition is evident.4 107. If the angle of intersection of the circles s, st be denoted by af/, it follows at once, from equation (13), that the mutual power (a/3) is equal to sin ra. sin rp cos aB. By this substitution, equation (405) is transformed into cos 11', cos 12', cos 13', cos 14', cos 15' cos 21', cos 22', cos 23', cos 24', cos 25' cos 31', cos 32', cos 33', cos 34', cos 35' = 0. (406) cos 41', cos 42', cos 43', cos 44', cos 45' cos 1', cos 52', cos 53', cos54', cos 55' 108. If the second system of circles coincide with the first we have, for any system of five circles on the sphere, 1, cos 12, cos 13, cos 14, cos 15 cos 21, 1, cos 23, cos 24, cos 25 cos 31, cos 32, 1, c, cos 34 35 =0 (407) cos 41, cos 42, cos 43, 1, cos 45 cos 51, cos 52, cos 53, cos 54, 1 * This theorem is the fundamental one in a Memoir by HIERR G. FROBENIUS, "Anwendungen der Determinantentheorie auf die Geometrie des Maasses." CRELLE'S Journal, Band 79, pp. 185-245, for the year 1875. It is also given in the Philosophical Transactions, vol. 177, part 2, for the year 1886, in a Memoir by R. LACHLIN, B.A., "On Systems of Circles and Spheres." Mutual Power of Two Circles. 115 Cor. 1.-The condition that four circles should be cut orthogonally by a fifth is 1, cos 12, cos 13, cos 14 cos 21, 1, cos 23, cos 24 = o0. (408) cos 31, cos 32, 1, cos 34 cos 41, cos 42, cos 43, 1 For in this case cos 15, cos 25, &c., vanish. Cor. 2.-The condition that four circles should be tangential to a fifth is 0, sin'2 12, sin2u 13, sin2 14 sin2-21, 0, sin2 l23 sin2 -24! =o. (409) sin2 331, sin22 32, 0, sin2 33 (409) sin2 41, sin2 42, sin2+43, O For if the circle S5 touch each of the circles s,, s2, s3, 84, cos 15, cos 25, &c., become each equal to unity, and subtracting each of the four first columns from the last, we get the result just written. 109. If t12 be the arc of a great circle which is the common tangent of two small circles whose spherical radii are ri, r2, and angle of intersection 132, then it may be proved by equation (13) that the mutual power of the two circles is equal to sin ri sin r2 - 2 cos rl cos r2 sin2 1 t2; and equating with the value sin r1 sin r2 cos q01 of ~ 107, we get sin2 123 = sin2 ~ t2. cot r1 cot r2. (410) Ience in the determinant (409) the sines of half the angles of intersection of the circles s8,,8, 83, s 8may be replaced by the sines 2 116 Small Circles on the Sphere. of half their common tangents, and denoting for shortness by 12 the sine of half the common tangent of the circles s, 82, the condition is 0, 122, 132, 14 -2 - — 2 2Fi O, 0 3, 24 -2 — 2, -2 = o (411) 3l, 32, 0, 34 2 1-=2 2 41, 422, 43, 0 which, expanded, is equal to the product of the four factors 12.34 ~ 23.14 + 31.24 = 0. (412) This theorem was first published in the Proceedings of the Royal Irish Academy, in a Paper by the author " On the Equations of Circles," in the year 1866. 110. If 8s, 82, 83, 84 be a system of four great circles, and 81s, 82, 83, 8s4 four other circles (great or small), then 11', 12', 13', 14' 21', 22', 23', 24' =0. (413) 31', 32', 33', 34' 41', 42', 43', 44' This is proved like Frobenius's theorem by multiplying the two determinants (xl, y2, z3, c r4), (x1', Y2', %'3 cos r4), the first of which vanishes; since rl, r2, r3, r4, being the spherical radii of great circles, are each equal to a quadrant, and their cosines vanish. 111. If the second system in ~ 110 be great circles, and coincide with the first, we get, since the mutual power of two great Mu-tual Power of Two Circles. 117 circles is equal to the cosine of the arc joining their poles, a relation identical with equation (408) for the six arcs joining four points on a sphere 12, &c., denoting in this case the arcs joining the points 1, 2, &c. If the three first points be the vertices A, B, C of a spherical triangle, and the fourth any arbitrary point D, whose distances from A, B, C are denoted by a, /, y, respectively, we get1, os, cb cos, cos b cos a cos c, 1, cos a, cos, =0. (414) cos b, cosa, 1, cos 1 c y cos a, cos /, cosy, 1 112. If the first three circles of the second system in ~ 110 coincide with the first three circles of the first system, and the poles of these circles be the angular points of a spherical triangle ABC. Also, if s4/ be a great circle distinct from s4, and the distances of the poles of these circles from the points A, B, C be a,,, y; a', 3', y', respectively, and 8 the arc joining their poles, we get1, cos c, cos b, cos a cos c, 1, ca cos a, cos = o. (415) cos b, cos a, 1, cos y cos a', cos /', cos y', cos EXERCISES.-XXX. 1. The incircles of a triangle and its colunar triangles have a fourth common tangential circle.-(HART.) For if a, b, c be the sides of the original triangle, the direct common tangents of the incircles of the colunar triangles are (b + c), (c + a), (a + b), respectively; and the transverse common tangents of the incircle of the 118 Small Circles on the Sphere. original triangle and incircles of colunar triangles are (b - c), (c- a), (a - b). Hence (see Art. 108) we have 23 = sin (b + c), 3 = sin (c+a), 12 = sin (a+b), 14 = sin (b + c), 24 = sin(c-a), 34 = sin (a +b). Hence the condition (412) is fulfilled. 2. Prove that the mutual power of two cireles is equal to the mutual power of two circles inverse to them. 3. If p, q, r be the normal co-ordinates of a point on the sphere, with respect to the sides of a spherical triangle ABC; prove that they are connected by the relation -1, cos C, cosB, p cos C, -1, cos A, q cos B, cos A, -1, r p, q, r, -1 In the equation (414), Art. 110, let the triangle ABC be replaced by its supplemental triangle, while the point D retains its position. 4. If a, b, c be the mutual distances of the spherical centres of three small circles whose radii are rl, r2, r3; prove that if r be the radius of a circle cutting them orthogonally, 4n2sec2r = 1, cos c, cos b, cos rL cos c, c1, cos a co r cos b, cos a, 1, cos rs cos ri, cos r2, cos r3, 0 CHAPTER VII. INVERSIONS. SECTION I.-INVERSION IN SPACE. 113. DEF. XXXVII.-Being given a fixed point S and a system ofpoints A, B, C..... if upon the right lines SA, SB, SC a system of points A', B', C'.... be determined by the relation SA. SA' = SB. SB' = SC. S C,.c. = constant, say k2, the two systems A, B, C... A'B'C'.... are said to be inverse of each other. The point S is called the centre of inversion, and the sphere whose centre is S and radius k, the sphere of inversion. 114. The figure inverse to a plane is a sphere passing through the centre of inversion. S A5XB A B Fig. 45. DEM.-From S draw the right line SA perpendicular to the plane P, and in P draw any line AB through A; then (Sequel, Prop. xx., p. 41), the inverse of the line AB is a circle SA'B' passing through S. Now, if the whole figure, consisting of the 120 Inversions. line AB and the circle SA'B', turn round the line SX, the line AB will describe the plane P, and the circle S4'B' will describe a sphere, which is the inverse of the plane. Cor. 1.-The inverse of a sphere passing through the centre of inversion is a plane. Cor. 2.-If m', n' be the inverses of the points m, n, then k2mtAn' mn= S' S(417) This follows from the triangles Smn, Sm'n', which are evidently similar. 115. The inverse of a sphere which does not pass through the centre of inversion is a sphere. DEM.-If X be any circle coplanar with S, its inverse will be another circle X', coplanar with S and X, and S will be the centre of similitude of the two circles (Sequel, Prop. I., p. 95); then, if the figure consisting of the two circles be turned round the line through the centres of both circles, the spheres described will be inverse to each other with respect to the point S. Cor. 1-The inverse of a circle with respect to any point in space is another circle. For the first circle may be regarded as the curve of intersection of two spheres; its inverse will be the curve of intersection of the inverse spheres. Cor. 2.-The cone which has for base a small circle of the sphere, and vertex any point, cuts the sphere again in another circle. EXERCISES.-XXXI. 1. The locus of a point, from which two unequal spheres can be inverted into two equal spheres, is a sphere. 2. The locus of a point, from which three unequal spheres can be inverted into three equal spheres, is a circle. Stereographic Projection. 121 3. DUPUIS' THEOREM.-If a variable sphere touch three fixed spheres, the locus of its point of contact with each fixed sphere is a circle. For if a variable sphere be inscribed in a trihedral angle, the locus of its point of contact on each face of the trihedral is a right line; and when we invert, the planes become spheres, and the right lines circles. 4. Prove that four unequal spheres can be inverted into four equal spheres. SECTION II.-STEREOGRAPHIC PROJECTION. 116. DEF. XXXVIII.-Stereographic projection is the drawing of the circles of the sphere upon the plane of one of its great circles (called the plane of the primitive) by Uines drawn from the pole of that great circle to all the points of the circles to be projected. It is evident that the plane is the projection in space of the sphere, the value of the constant k2 being 2R2. 117. The stereographic projection of any circle is a circle. This follows at once from ~ 115, Cor. 1, but we here give an independent proof. DEM.-l1. In the case of a small circle. Let Pm be a generator of the cone, touching the sphere along p P C o Fig. 46. the given circle, and let P', m' be the projections of the summit 122 Inversions. of the cone, and of the point m of the circle, on the plane of the primitive. Let O be the pole of the primitive, and let Pm produced meet a tangent plane to the sphere at O in T; then, since the plane of the primitive and the tangent plane at O are parallel, the plane OmP cuts them in parallel lines (Euc. XI., xvI.). Hence the angle P'm'O is equal to m' OT; but the angle m'OT is equal to OmT, since the tangents mT, OT are equal. Hence the angle P'm'O is equal to the supplement of Pm 0, and the angle O is common to the two triangles PmO, P'm'O; therefore OP: Pm:: OP': P'm'; and since the three first terms of this proportion are given, the fourth, P'm', is given. Hlence the locus of m' is a circle whose centre is collinear with the pole of the primitive and the vertex of the cone.-(CHASLEs.) Cor.-If the cones circumscribed to a sphere along a system of circles have their vertices in a line passing through the pole of the primitive, their stereographic projections is a system of concentric circles. 2~. In the case of a great circle. Let A be the pole of the primitive, and let the plane of the circle AIB be perpendicular to the line of intersection of the I CD Fig. 47. plane of the primitive with the plane of the circle to be projected, and let it intersect the plane of that circle in the line Stereographic Projection. 123 EF. Let IC be perpendicular to AB. Join AE, AF, intersecting IC in the points G, IT. Now, since GAH is a right angle, and A C is perpendicular to GEI, the angle AHG = GA C (Euc. VI. viII.) = AEC (Eue. I. v.). Hence the triangles EAF and HA G are inversely similar, and therefore the section made by the plane of the primitive with the cone, whose vertex is A, and which stands on the great circle EF, is an antiparallel section. Hence it is a circle. Cor. 1.-The projections of the poles P, Q of the great circle EFwill be inverse points with respect to its projection. Cor. 2.-If the plane of a small circle be parallel to the plane of EF, the projection of P and Q will be inverse points with respect to its projection. Cor. 3.-A system of small circles, whose planes are parallel, will project into a system of coaxal circles. Cor. 4.-Every circle whose plane passes through the pole of the primitive is projected into a right line. 118. The angle made by any two circles on the sphere is equal to the angle made by their projections on the plane of the primitive. DEM.-Let O be the pole of the primitive, M2 the point in which the circles intersect; and let MT, MFV the tangents to the circles at Jf meet the tangent plane to the sphere at O in the points T, V. Join OT, 0V, TV; then evidently the angle 2T'V = TOV; but since the tangent plane at O is parallel to the plane of the primitive, the lines OT, 0 V are parallel to the projections of the lines MT, MV. Hence the angle TO Vis equal to the angle between the projections. Cor. 1. -Any circle whose plane is perpendicular to the plane of the primitive is projected into a circle orthogonal to the primitive. Cor. 2.-A system of coaxal circles on the sphere is projected into a system of coaxal circles on the plane of the primitive. 124 Inversions. For a system of coaxal circles on the sphere is intersected orthogonally by a system of circles passing through the two limiting points (~ 94). Hlence the projections are intersected orthogonally by a system of circles passing through two points. 119. Applications to Spherical Trigonometry. Let ABC be a spherical triangle, AB'C a colunar triangle; then, if 0, the pole of the primitive, be the antipodes of A, the sides AB, A C, AB' will project into right lines ab, ac, ab', and the circle BCB' into the circle bcb'. Join bc, cb'; then the angles of the figure formed by the lines ab, ac, and the arc bc, are respectively equal to the angles of the spherical triangle ABC (Art. 117); but the sum of the angles of the rectilineal triangle abc is two right angles, hence the sum of the angles formed by the arc bc with its chord is equal to the spherical excess 2E; therefore one of them is equal to E. A OFig. 4 Fig. 48. Or thus:-If a circle be described about the triangle abc, the angle made by this circle with the arc bo is (~ 118) equal to the angle made by the circumcircle of the triangle ABC with the side BC, and this is equal to (A - E) (Exercises xxIIi. 15); Stereographic Projection. 125 and the angle made by the circumcircle of abc with the chord bc is equal to A (Euc.. III. XXI.). Hence the angle between the arc bc and its chord is equal to E. Cor. 1.-If the radius of the sphere be unity, the sides of the rectilineal triangle can be expressed in terms of the spherical triangle. Thus, evidently, ab = tan A OB = tan - AB = tan c. (418) ac = tan AOC= tan A C = tan b. (419) Again, bc: ab: sin bac: sin acb: sin A: sin (C-E);:n n bc: tan-c:: n:.; sin b. sinc 2 sin a. sin - b. cos O c bc= sin I a (420) cos - b cos - c Similarly, b'c cos i. (421) cos b sin mc The equation (420) may be got from equation (417) by putting k2 = 2. Thus2 chord B C sin a bc = OB. OC cosi b cos c' and (421) from (420), by the substitution of ~ 78. Cor. 2.-If the angles of the rectilineal triangle abc be denoted by a, fi, y, we have a=.A, /f =B-E, y= C-E. (422) ExERCISES.-XXXII. 1. Prove the fundamental formula (13) by stereographic projection. From the triangle abc we have (bc)2 = (ca)2 + (ab)2 - 2 (ca) (ab) cosdA, and substitute from equations (418)-(420). 2. Prove Napier's Analogies. tWehave an ( - ) ac - ab We have av t -eY) tan ~ a =ac + ab' And substitute frorn equations (418)-(422). 126 Inversions. 3. Prove Delambre's Analogies. From ab e = (ac + ab) sin. a (ac - ab) cos a From abc we have bc = = cos ~ (j - y) sin ~ (8 - y) and substitute as before. 4. Being given three circles on the sphere; there are eight points on the sphere, any one of which, if taken as the pole of the primitive, the three circles will be projected into three equal circles.-(STEINER.) 5. tan 2 = Vtan 2 s. tan ~ (s- a) tan (s - b) tan (s - c). Express tan ~ the angle ab'c, in terms of the sides of ab'c. 1 + cos a + cos b + cos c 6. Prove that cos E = 4 cos 2 a cos ~ b cos ~ c In the triangle ab'c, we have ab' = ac cos cab + cb' cos cb'a. 7. To express the spherical excess of a spherical quadrilateral in terms of its sides and diagonals. c' u t\\ / c Fig. 49. Let ABC:D be the quadrilateral; and denoting the sides and diagonals AB, BC, CD, DDA, AC, BD by a, b, c, d, e, f, respectively; then taking the antipodes of A for the pole of the primitive, the arcs AB, AD, AC will project into right lines ab, ac, ad; and the arcs BC, CD into arcs of circles Steregraphic Projection. 127 bc, cd; then drawing bt, ct tangents to bc, and cu du, tangents to cd, we have in the plane hexagon abtcud the sum of the angles A + + B+C+ D+t+u=47r; but A+ + B -C D-)=2r+22E;. 2E =27r-t-u. Again, if the circles bc, cd intersect again in c', c' is the stereographic projection of the antipodes. Hence the right line ca produced will pass through c. Join bc, bc'; dc, dc', then the angle tbc = bc'c. Hence bc'c is half the supplement of t, and cc'd half the supplement of u;.'. 2bc'd+t + u=2r;.'. bc'd =E. Now from the plane triangle bc'd, we have in 'd = sin2 1 = (b' + bd - c'd)(bd + c'd - bc') 4bc'. dc' but sin f cos ~b cos b bd = osi c d= s cos I a cos cos da COS, sin c' cos d sin e.sin2 E = (sin le. sin f + cos a cos c - cos cos csd) (sin e sin f - cos 1a cos c + cos b cos d} 4 cos a cos b cos c cos d 8. If a spherical quadrilateral be cyclic, prove that. in2 1 = -æ)in (,( s - b)a) sin (s-b) sin (s - c)sin (s-d) (4-) sin" E = 2ff 2,424 a b c d * cos cos - cos - cos - 2 2 2 2 9. If the cyclic quadrilateral be circumscribed to another circle, prove sin2 E = tan a. tan b tan c. tan2 d. (425) 10. Being given four circles in a plane, prove that the plane can be inverted into a sphere, so that the four circles on the plane will be the stereographic projections of four equal circles on the sphere.-(STEINER.) 11. If A', B', C' be the stereographic projections of the angular points of the spherical triangle ABC; and if the angles of the plane triangle A'B'C' be respectively equal to those of the spherical triangle, each diminished by one-third of the spherical excess, prove that the arcs drawn from A, B, C to one of the poles of the primitive divides the area of ABC into three equal parts. Observation.-The applications of Stereographic Projection to Spherical Trigonometry, contained in ~ 119 and in Exercises xxxii., are taken from M. PAUL SERRET, Méthodes des Géométrie, pp. 30-44. CHAPTER VIII. POLYHEDRA. SECTION I.-REGULAR POLYHEDRA. 120. If S be the number of solid angles, F the number of faces, E the number of edges of any polyhedron (see Appendix to EUCLID, 6th edition, p. 283), 8 + F=E + 2.-(EULER.) DEM.-With any point in the interior of the polyhedron as centre, describe a sphere of radius r, and draw lines from the centre to each solid angle; let the points in which these meet the surface of the sphere be joined by arcs of great circles. These arcs will divide the surface into F spherical polygons. Now if s denote the sum of the angles of any of these polygons, and m the number of its sides, its area is r2(s - (m - 2)7r), but the sum of the areas of all the polygons is equal to the surface of the sphere or 47rr2. Hence, since there are F polygons, we have 47r = ss - 7rm + 2F7r; but Ss is evidently equal to 27rS, and 2m is the number of the sides of all the polygons, and therefore equal to 2E. Ience 47r = 27rS - 2Er + 2F7r;.-. S+_F=E+2. (426) 121. There can be onlyfive regular polyhedra. DDEM.-Let m be the number of sides in each face, and n the number of plane angles in each solid angle, then the entire number of plane angles is equal to mF or nS or 2E. Hence we have the equations mF=nS=2E and S+F=E+2. Therefore solving for S, F, and L, we get 4m _4n 2mn B -.F li=. E= 2(m+n)-mn' 2(m+ n)-mn' 2(m+n)-mn (427) Regular Polyhedra. 129 Since the denominator in these expressions must be positive, — + - must be greater than;but n cannot be iess than 3, mn n 2i since a solid angle cannot be formed by less than three plane angles. Hence m cannot be greater than 5. The following will be found to be the only admissible system of values for m and n, viz., 3, 3; 4, 3; 3,4; 5, 3; 3,5; and the corresponding polyhedra are the Tetrahedron, Cube, Octahedron, Dodecahedron, and Icosahedron, or solids of 4, 6, 8, 12, 20 faces, which have respectively 4, 8, 6, 20, 12 vertices. 122. If I denote the inclination of two adjacent faces of a regular polyhedron, 7r 7r sin I = cosec -. cos -. (428) m n B D Fig. 50. DEM.-Let AB be the side common to the two faces, C and D their centres, from which let CE, DE be drawn perpendicular to AB; then the angle between CE and ED will be equal to I. In the plane of the lines CE, DE, let CO and DO be drawn at right angles to them, and meeting in O. Join O0A, OE, OB, and from O as centre suppose a sphere to be E 130 Polyhedra. described, cutting 0A, OE, OC in the points a, e, c, respectively; then aec is a spherical triangle, having the angle e right; 77 7r also cae = and ace = -, and by equation (111), sin ace = cos cae n m cos ce; but cos ce = cos coe = sin I; 7r 7r. sin I= cosec- cos -. m n Cor. 1.-The following are the values of Ifor the five regular polyhedra. Thus, denoting them by P4, P6, P8, P12, P20:1 7r i In P4, cos=-; in P6, I= in P8, cos I= - - 3 2; 3' in P2, cos I-; in P20, cos = - /5. -/5 3 Cor. 2.-If r be the radius of the inscribed sphere, and a a side of one of the faces, a 7T i r= cot-. tan. (429) 2 in 2' I a r I For r = CE. tan CEO = CEtan cot - tan 2 2 m 2 Cor. 3.-If R be the circumradius of the polyhedron, a ir I R= - tan- tan-. (430) 2 n 2 Cor. 4.-The surface of a regular polyhedron ma2 F 77 = maFcot -. (431) 4 m Cor. 5.-The volume of a regular polyhedron ma2 rF 7r = 12 cot (432) 12 c Cor. 6.-The octahedron is the reciprocal of the cube, and the icosahedron of the dodecahedron. Parallelopipeds and Tetrahedra. 131 EXERCISES.-XXXIII. 1. Find the values of E, F, S for each of the regular polyhedra. 2. Prove that the centres of the faces of the polyhedra P4, P6, P, P2, P, 20 are respectively the summits of polyhedra P4, P8, P6, P20, P12. 3. Find the ratios between the volumes of a tetrahedron or cube and the volume of the solid, whose summits are the centres of its faces. 4. Prove that the inradius of P4 = thee times its circumradius. 5. In the same' case, the radius of the sphere touching its six edges is a mean proportional between the inradius and circumradius. 6. Prove that the ratio of the inradius to circumradius is the same in P6 and Ps, and also in P12 and 2Po. 7. In any convex polyhedron (regular or irregular), prove that the number of faces having an odd number of sides is even, and that the number of solid angles having an odd number of edges is uneven. 8. In every convex polyhedron, the number of triangular faces increased by the number of trihedral angles is equal to or greater than eight. 9. Every convex polyhedron must have either triangular, or quadrangular, or pentagonal faces, and trihedral, or tetrahedral, or pentrahedral angles.* SECTION II.-PARALLELOPIPEDS AND TETRAHEDRA. 123. To find the volume of a parallelopiped in terms of three conterminous edges and their inclinations. Let OA, OB, OC be the three edges, and let their lengths be a, b, c, respectively; and let the angles BOC, COA, A OB be denoted by a, /3, y. Draw AD perpendicular to the plane B O C, and describe a sphere, with O as centre, meeting the lines OA, OB, OC, OD in the points a, b, c, d, respectively. The * The most important recent works which treat of the polyhedra are ALLMAN'S "Greek Geometry from THALES to EUCLID," and "Lectures on the Icosahedron," by PROFESSOR KLEIN, Gottingen. This is a very remarkable work, showing the great importance of the polyhedra in the higher departments of modern Analysis. R 2 132 Polyhedra. volume of the parallelopiped is equal to the product of the base and altitude = bc sin a. AD1; but A1) = a. sin A 0D = a. sin aod;. vol. = abc sin a. sin aod = 2abc. n, (433) n being the first staudtian of the solid angle O - ABC = abc \/1 - cos2a - cos2/ - cos2y + 2 cos a cos 3 cos y. (434) A Fig. 51. Cor. 1.-The volume of the tetrahedron O - ABC = abecn. (435) Cor. 2.-To find the volume of a tetrahedron in terms of its six edges. Let BC a', CA = b', AB = c'; then we have b2+ - a'2 + a2 - b' a2 + b2- -2 sa= 2bc 2 cos = 2c= ab = abc v/ 1 - cos2a - cos2f - cos3y + 2 cos a cos p cos y. Hence 144 F2 = Sa2a'2(b2 + b'2 2 + ' + ' a2 - a'2) - a2b'2 '2- b2 '2a'2- 2a'2b'2- a2 b2 2. (436) Cor, 3.-If T~, Tb, T, be the areas of the three faces OBC, OCA, OAB of the tetrahedron O - ABC, and N the second staudtian of the solid angle O-ABC, V2 = T T,bTcN. (437) Parallelopipeds and Tetrahedra. 133 For TI - bc sin a, Tb = ca sin, Tc - ab siny, 2n2 sin a. sin/ P. sin y Hence I T2aTbTYN = - a2 b 2 n2 = V2. Cor. 4.-The second staudtians of the solid angles of a tetrahedron are proportional to the areas of the opposite faces. This follows at once from Cor. 3. Cor. 5.-The volume of a tetrahedron is equal to î of the product of the areas of two faces by the sine of their dihedral angle divided by the length of their common edge. For the vol. = i of the triangle OB C. AD, and AD = 2 triangle AB C, multiplied by sine of the dihedral angle of the faces OB C, AB C divided by B C. Cor. 6. —The products of opposite edges of a tetrahedron are proportional to the products of the sines of the corresponding dihedral angles. 124. If a, /3, y, 8 denote the areas of the four faces of a tetrahedron, a2 = j2 + y2 + 8 + 2/,y cos (/3y) + 2y8 cos (y8) + 283 cos (88). (438) DEM.-We have a = 8 cos (a/)) + y cos (ay) + 8 cos (a8). (1) 3= a cos (Ba) + y cos (/ly) + 8 cos (,/8). (2) y= a cos (ya) + ( cos (y,/) + 8 cos (y78). (3) 8 a = cos (8a) + B cos (8) + y cos (8y). (4) Multiplying by a, 3, y, 8, respectively, and subtracting the sum of the three last products from the first, we get the above result. 134 Polyhedra. Cor.-If we eliminate a, /3, y, 8 from the equations (1), (2), (3), (4), we get the following relation between the six dihedral angles of a tetrahedron: — -1, c, cos ay, cos a8 cos/ a, -1, cos /y, cos 8 cos ya, os y/, - 1, cos ya I (439) cos 8a, cos 8,, cos Sy, -1 This relation may also be easily inferred from equation (414.) 125. Tofind the diagonal of a parallelopiped in terms of three conterminous edges and their inclinations. Fig. 52. B E Fig. 52. Let the edges OA, OB, OC be denoted by a, b, c; and the angles OBC, OCA, OAB by a, /, y, respectively; let OD be the diagonal required, and OE the diagonal of the face OAB; then OD2 = OE2 + EL+ 20E. ED. cos COE = a2 + b2 + 2ab cos y + 2 + 2c. OE. cos COE. Describe a sphere, with O as centre, cutting OA, OB, OC, OE Parallelopipeds and Tetrahedra. 135 in the points a, b, c, e, respectively; then we have, by Stewart's theorem, cos COE = (cos a. sin a Oe + cos f sin b Oe) sin y; 0. 2 = a2 b2+ c2 + 2ab cosy 2e. 0.E + O (cos a. sin a Oe + cos,8 sin b e); sin y but OE sin a Oe = b sin y, and OE. sin b Oe = a sin y. Eence D2 = a2+ b2 + 2 + 2bc cos a + 2ca cos, + 2ab cos y. (440) 126. To find the radius of a sphere circumscribed to a tetrahedron. D AF. --- Fig. 53. ErrST METHOD (STAIDT'S):Through the centre O of the circumsphere draw a plane perpendicular to the radius DO, cutting the sides of the trihedral angle D in the plane triangle A'B'C'. This plane being parallel to the tangent plane to the sphere at -D, Â'B' is antiparallel to AB in the angle ADB, B' C' BC,, BDC, C'A', CA, CDOA. 136 Polyhedra. Let DA = a, DB = b, DC= c, BC = a', CA = b', AB = O', DA'= a, DB' = S, DC = y, B' C= a', C'A'= 3', A'B'= y', we have a' _3 _y, _ a' a'/3b a'.aa a a= - = - = =aa.; a' c b c bc bc abc /3 = bb*. b Y= cc'. abc abcC If O' be the second extremity of the diameter DO' of the sphere, we have aa = b/3 = cy = 2R2 = DO. D0'. aa'. 2R2 bb'. 2R2 cc'. 2R2 Hence a' =, ---, ' y= abc abc abc Putting aa' = a, bb'= b,, cc' = ci, and al + bl + cl = 21s, we have triangle 4R4 4R4 'B' C' = 222 s. (sl - al) (sl - bi) (sl - Ci) = a s2.2 We have also DAB C abc a2 b'2 a2 b2c2 DA'B'C' afly aa. b,8. cy 8R6 ' then DABC a2b2c2,S1 4R4 =;.' R = 6. DABC- (441) a2b2,02 SECOND METHOD (DOSTOR, Nouvelles Annales, 1874, p. 523):Let DABC be the tetrahedron, O its circumcentre; A', B', C' the middle points of the edges DA, DB, DC, which, as before, are denoted by a, b, c, respectively. Draw 0f parallel to DA, meeting the face BDC in I, and M1N parallel to BD. Now, Parallelopipeds and Tetrahedra. 137 since the projection of DO, and of 2D, NM, MO on DA, DB, DC, DO are equal, we have i a = DNV cos ac + N-M cos ab + MO. - b = DN cos bc + NM[ + JIO cos ab, c = DN + NM cos bc + MO cos ac, R = DN cos (c, R) + NXfcos (b, R) + MO cos (a, R). D / ^ ^N A C B Fig. 54. Hence 2R21 MO. a + J31N. b + 2D2. c. a = 2M0 + 2MXN cos ab + 2DN cos ac. b = 21I0 cos ab + 2MN + 2DY cos bc. c = 2M0 cos ca + 2MXV cos cb + 2DN. Hence, eliminating MO, MN2, DN, we get 2R2, a, b, c a, 2, 2 cos ab, 2 cos ac =0. h, 2 cos a, 2, 2 cos bc c, 2 cos ca, 2 cos eb, 2 138 Polyhedra. Now'n being the first staudtian of the solid angle D- ABC, we have 4n2= 1, cos ab, cos ac cos ba, 1, cos bc. cos ca, cos bc, 1 Therefore 64Rn2 = O, a, b, c 0, a2, b2, c2 a, 2, 2 cosab, 2cosac 1 a2, 2a2, 2abcosab, 2accosac b, 2 cosba, 2, 2 cos bc a22 b2, 2bacosba, 2b2, 2bccosbc c, 2 cos a, 2 cosb, 2 2, 2 a cos a, 2 cb cos b, 2c2 Hence 64a2b2C2n2.R2 = 0, a2, b2, C2 a2, 0) c2, b/2 = - 1 (aa')4+ 24 (aba'b'). (442) b2, c'2, O, a'2 C2, b'2, a'2 0 Cor.-24 VR = { 2s (aba'b') - S (aa')4. 127. The Isosceles Tetrahedron.-(NEUIBERG.) DEF. XXXIX. —An isosceles tetrahedron is one whose opposite edges are equal. From the definition it follows at once (Euc. I., viI. xxxmi.) that the four faces are equal, and that the sum of the plane angles forming each trihedral angle is equal to two right angles. 128. If we suppose BC= AD =a, A C= DB=p, AB=DC=y; then denoting by a, b, c the angles of the triangle ABC, they are also the face angles of the trihedral angle D - ABC; and Parallelopipeds and Tetrahedra. 139 representing by A, B, C the dihedrals DA, DB, DC, we have (~ 29)1~. sina: sinb: sin:: sinA: sinB: sin C:: a: /: y. (443) 2~. The first staudtian of D- ABC = v /i - cos2a - cos2b - cos2C + 2 cos a cos b cos C = </cos a cos b cos c. (444) 3~. If MJi N, P, Q, R, S be the middle points of the six edges, we have PQ = AC= BD = PN; then PQMN is a lozenge, and MP is perpendicular to NQ. Ience the three medians XP, NQ, RS form a system of three rectangular axes. D M N.. Fig. 55. 4~. Since the tetrahedron D)AB C can coincide with ADCB, BCDA, CBAD, the four lines DG, AG, BG, CG are equal. fence G, the centre of gravity of AB CD, is also the centre of the circumscribed sphere, and of the inscribed sphere. 5~. The inscribed sphere touches the faces at the centres of the circumcircles. 140 Polyhedra. 6~. The medians PMQ, QN, RS are perpendicular to their corresponding edges. 129. The four solid angles G- AB C, G- BD, -, G-BCD, G -A CD, are equal. Hence the spherical triangles which they intercept on the sphere are each equal in area to one-fourth of the spherical surface, and therefore the spherical excess of each is two right angles. Again, the angle JlGg = supplement of iXGD = supplement of MGC = PGC = A GB. Then, in the spherical triangle ABC, the arc which joins a summit to the middle of the opposite side is equal to the supplement of haZf that side, and the arcjoining the point of concourse of the medians to the middle of any side is equal to half that side. HIence the spherical triangle ABC is divided by the antipodes of the point D into three diametral triangles. 130. The volume of the tetrahedron is double of the octahedron MINPQRS = 16SMNG = 8 GM. GN2. GS; but GS2 + GM2 iBC2C2= G2 + GN2 32, GN2+ GS2= 72; 9. F- /(2+7 a2 2 _ )(y2+-2 _ 2) (a2+2 2) (445) Cor. 1.-The square of the radius of the circumscribed sphere a2 + 32 + y2 8= a +, + 2 (446) For A G2= AfZM + ffG2 = 2+ 22 2+ C8 -- 8 3V Cor. 2.-The radius of the inscribed sphere = AB (447) ja.Jj( Parallelopipeds and Tetrahedra. 141 EXERCISES XXXIV. 1. If the four edges of a tetrahedron be tangents to a sphere, the sum of each pair of opposite edges is constant. For if ti, t2, t3, t4 be the tangents drawn to the sphere front the vertices of the tetrahedron, it is evident that the sum = tl + t2 + t3 + t4. 2. If a, a' be two opposite edges of a tetrahedron, and d their shortest distance, the volume 1 A = aa'd sin (aa'). (448) 3. The four escribed spheres of an isosceles tetrahedron are equal, and the radius of each is equal to the diameter of the inscribed sphere. 2t, t2 ta t4 4. Prove that the radius of the sphere in Ex. 1 = 23 t. (449) 5. If V be the volume of a tetrahedron, whose edges of a face are a, b, c, and opposite edges a', b', c'; and V' the volume of a tetrahedron, whose edges of a face are a', b', c', and opposite edges a, b, c; then 144 ( v2 - Y'2) = (a2 - a'2) (b2 - b'2) (c2 - c'2). (460) (WOLSTENHOLME.) 6. If (a, a'), (b, b'), (c, c') be the three pairs of opposite edges of a tetrahedron, and denoting by the same letters the dihedral angles adjacent to these edges, prove that if the altitudes cointersect1~. a2+a2 = b2+b'2=c2 + c2. (451) 2~. cos a cos a' = cos b cos b' = cos c cos c'. (452) 7. If the lines joining the summits of a tetrahedron to the points of contact of opposite faces with the inscribed sphere cointersect, prove that cos ~ a cos a' cos b. cos b' = cos c. cos c'. (453) 8. If the spherical triangles which are equivalent to the two trihedrals D - ABC, G - ABC (fig. Art. 127), be denoted by ABC, A'B'C', respectively, prove tan a' inA (454) / sin b sin c cos a 142 Polyhedra. 9. If ABCD be a tetrahedron, and if we denote by AB the angle between the faces ABC, ABD, prove that ~4-~2.C/2 sn2 AtD- A 1 AB2. CD. sin2 (AB. CD) = ABC2 + ABD2 - 2ABC. ABD cos AB, where ABC denotes the area of the triangle ABCe (455) Project the triangle BCD into B'C'D' on a plane perpendicular to AB; we have then A C'D' = CD sin (AB CD), and C'D'2 = B'C'2 + B'D'2 - 2B'C'. B'D'. cos C'B'D'; D C' BFig. 56 Fig. 56. then, multiplying by AB2, and remembering that B'C', B'D' are equal to the altitudes of the triangles ABC, ABD, the proposition is proved. 10. If MD be any point in the edge CD, prove that A ABM2. CD2 = ABC2. MD2+ ABD2. CM2 + 2ABC. ABD. CM. MD cosAB. (456) Draw M'Pparallel to BD, and we have B'M'2 = B'P2 + M'P2 + 2B'P. PM' cos AB; B'P CHM PM ' 7MD also FB') = C' B'C' = - &c. 11. If M be the middle point of CD, AB2 + ABD2 +. ABD cos A. (457) 4ABM2= ABC2 + ABD2 + 2ABC..ABD. cos AB. (457) CHAPTER IX. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO GEODESY AND ASTRONOMY. SECTION I.-GEODESY. 131. To reduce an angle to the horizon1~. General Solution.-Let OZ be the vertical of the observer at O; then, if the angle MOON= a, NOZ = b, MOZ = c, it is required to find the projection of MON on a horizontal plane passing through 0. 2 N A M Fig. 57. Sol.-From 0 as centre, with a unit radius, de scribe a sphere, cutting OZ, OM1, ON in A, B, C, respectively; then the sought angle is the measure of the dihedral angle BOA C, or of the angle A of the spherical triangle BA C, which is given by the formula tan- - A_ sin(s - b) sin(s - c) a sin s. sin (s - a) 144 Applications of Spherical Trigonometry. 2~. Solution of Legendre.-This solution;s applicable only when the angles of elevation of the objects M1, N are very small; that is, when b and c are each near 90~. It depends on the following lemma: —Being given a spherical triangle ABC, the angle A1 of the rectilineal triangle formed by its chords (called the chordal triangle) is given by the equation cos A, = sin b sin - c + cos b cos c cos A. (458) A B o Fig. 58. DEiM.-Let A'BC be the colunar triangle, M, N the middle points of the arcs A'B, A'C; then thc chords AB, C are parallel to the radii 0f, ON of the sphere. Hence cos A1 = cos MON= sin b sin - c + cos 1 b cos ] c cos 4. Cor.-If A1 = A - 0, then cos A4 = cos A + 0 sin A approximately; and substituting in (458) for cos 2b cos 1 c, sin i-b sin ac, the values cos2 (b+c) - sin2 (b - c), sin24 (b +)- sin2 1 (b - c), we get, after an easy reduction, 0 = tan - sin2 (b+ c) - cot 2 4 sin.2 (b - c). (459) Given the oblique angle contained between two objects above the horizon, tofind the corresponding horizontal angle. Geodesy. 145 Let A be the place of the observer, i, N the objects above the horizon; let a sphere of radius unity be described, touching the horizon at A, and intersecting the lines AM, AN in the points B, C. Draw the gréat circles ABO, ACO; then if Z, H' denote the elevations of A-M, AN above the horizon, we have Rf= - arc AB, H' = ar A C; that is, considering the spherical triangle AB C, if= c, H' = % b. Now the angle A of the spherical triangle ABC is the horizontal angle which corresponds to the oblique angle MXAN. Hence, if 0 denote the difference, we have (459) 0 = tan 'A sin2 (b + c) - cot A- sin2 ~ (b - c); 0 A Fig. 59..-. 0 = tan }i sins a (H + H') - cot -4 sin2 j (H - ( l). (460) In practice H, I' and 0 are very small. Hence this formula may be replaced by the following, which is Legendre's: 0= {i(f+ r')}2 tan J fMAN- ( (Hf- H')} cot IMAVN, (461) an approximate value of the difference between the circular measures of the oblique and horizontal angles, which must be added to the former to obtain the latter. L 146 Applications of Spherical Trigonometry. 132. Legendre's Theorem.If the sides of a spherical triangle be small compared with the radius of the sphere, and if a plane triangle be constructed whose sides are equal in length to those of the spherical triangle, then each angle of the spherical triangle exceeds the corresponding angle of the plane triangle by one-third of the spherical excess. DEM.-Let a, b, c be the lengths of the sides of the spherical triangle, r the radius of the sphere, then the circular measures of the sides are a b c r' r? r' respectively; hence a b c cos - - cos-. cos - Cr r r cos A =. b. c sin-. snr r a: 6 and, substituting for cos-, cos -, &c., their values given in Pl. Trig., ~ 158, we get, neglecting powers higher than the fourth of 1 r' cos = b2t — l-Ar2r2 24r- 2r2+ 24r4J 1 — 2r2C 24r4r/ r2 6r2 6r (b2c-i-2-'a2 a4- 64-c - 6b22\ / b2 + c2\ 2bc 24bcr2 ) 6r2 (b2+c2-a2 a4-b-c4_6^ C2) b2 + C2 = + 1 + 6r2 ~ 2bc 24bcr2 6r2 b2 + c2 _ a2 a4 + 4 C4 + 2(a2b2+ b22+ a2) 2bc 246cr2 Geodesy. 147 Hence, if a, /3, y denote the angles of the plane triangle, whose sides are a, b, c, we have bc sin2a cos A = cos a - 62, nearly. 6r2, e Now, putting. = a + 0, we have cos A = cos a - 0 sin a. b c sin a S tHence 6= -, 6rP = 3r2' S denotingTthe area of the plane triangle; 3r2' * -aSimilarly, B -B3rp ' -= S 3r2. Hence - = spherical excess;.'. A - a = spherical excess. (462) 133. The area of the spherical triangle is approximately equal to <1+ b2 + 2) (463) DEM._ 1 8u, s-a s-b s-c tan E= tan. tan.tan. tan 2; 2s2 s 2r\ 24r2' + but tan = 2r 1 2r 2r 2 8r2-. HIence Is s-a s-b s-c( s2 (s -a)2 a 8r2 + 2 12r21 + 12r2 L 2 148 Applications of Spherical Trigonometry. therefore E= S 1+ ( a)2+ ( - b)2+ (s ) 4r- + 12r2 Hence 2Er2= + 2 +2. Cor.-The area of the spherical triangle is equal to the area of the plane triangle, if we omit terms of the second degree. 1 in-. r 134. If n denote the number of seconds in the spherical excess, A the area of the spherical triangle on the surface of the earth in square feet; then log n = log A - 9-3267737.(GENERAL ROY.) DEM.-We have 2E= 0626; A = 206265 Now the mean length of a degree = 365155 feet. Thus - = 365155; 180 substituting the value of r from this equation in the value of A, and taking logarithms, we get log n = log A - 9'3267737. (464) EXERCISES.-XXXV. 1. The angles subtended by the sides of a spherical triangle at the pole of its circumcircle are respectively double of the corresponding angles of its chordal triangle. 2. Prove Legendre's theorem from either of the formulae for sin A, cos, A, tan ~ A, respectively, in terms of the sides. 3. If the radius of the earth be 4000 miles, what is the area of a spherical triangle whose spherical excess is 1~. 4. If A", B", C" be the chordal angles of the polar triangle of ABC, prove cos A" = sin ~ A cos (s - a), &c. (465) 5. If A'BC be the colunar of ABG; prove that the cosines of the angles of its chordal triangle are respectively equal to cos a cos, sinbsin (C-E), sin c sin (B - E) (466) Astronomy. 149 6. If B be the circumradius of a spherical triangle, Ai, Bi, Ci the angles of its chordal triangle; prove sin A = sin a cosec, sin = sin cosec sin Bi = R s in C i c cosec 2. (467) 7. Prove sin Ai: sin (B - C):: sin A: sin (B - C). (468) 8. Prove the proposition of ~ 132 from equation (351). 9. Prove E = (tan ~ a tan b) sin C - ~ (tan ~ a tan 2 b)2 sin 2C. (469) [Make use of the value of tan E drawn from equations (351), (356).] 10. Show that in every case of the solution of spherical triangles, except where the three angles are given, that Legendre's theorem may be used for an approximate solution. SECTION II. —ASTRONOMY. 135. Astronomical Definitions. If PHNR represent the meridian of any place, produced to meet the celestial sphere, P the north pole, O the south pole of z H / Fig. 60. the heavens, HR the horizon, EQ the equator, Z the zenith; then, for a place whose zenith is Z, QZ is the latitude; and 150 Applications of Spherical Trigonometry. since QZis evidently equal to PR, PR is equal to the latitude; but PR is the elevation of the pole above the horizon. Irence the elevation of the pole above the horizon is equal to the latitude. Again, if S be any heavenly body, such as the sun or a star, its position is denoted by any one of four systems of spherical co-ordinates as follows: 1~. The great circle ZST passing through the zenith and S, and meeting the horizon in T, is called the vertical circle of S. The arc HT, measured from the south point of the horizon, or its equal the angle 1ZT, is called the azimuth, and ST the altitude. lIT, ST are the spherical co-ordinates of the star S; ZS is its zenith distance, and the arc RT its azimuth from the north. 2~. Join SP, and produce to meet the equator in K. The arcs QK, XS form the second system of spherical co-ordinates; QK, or its equal the angle ZPS, is called the hour angle of S, and KS the declination. The declination is positive when S is north of the equator, and negative when south. The great circle PSK is called the declination circle, and PS the polar distance of S. 3~. The great circle which the centre of the sun, seen from the centre of the earth, appears to describe annually among the stars is called the ecliptic; and its inclination to the equator, which is nearly 23"~, the obliquity of the ecliptic. The points of intersection of equator and ecliptic are called the equinoxesone the vernal equinox (called also the first point of Aries), and the other the autumnal equinox (the first point of Libra). If wr denote the first point of Aries, then r'K is called the right ascension, and KS the declination of the star; crK, KS are the third system of spherical co-ordinates of S. The right ascension is counted eastward, from 0 to 360~. 4~. From S draw a great circle Sa- perpendicular to the ecliptic; then rqp-, aS are the fourth system of spherical coordinates of S, and are called respectively its longitude and Astronomy. 151 latitude. The longitude is reckoned eastward, from 0 to 360~. The latitude is positive when north, and negative when south. 136. If the small circle, M, M', passing through S, and parallel to the equator, represent the apparent diurnal motion of the sun or other heavenly body (the declination being supposed constant), it is evident he will be rising or setting at A (according as the eastern or the western hemisphere is represented by the diagram). He will be east or west at to, will be at B at 6 o'clock, morning or evening, will be at noon at M, and at midnight at MI'. 137. The foregoing definitions and diagram will enable us to solve several astronomical problems of an elementary character, such as the following:1~..Tofind the time of rising or setting of a known body. Consider the spherical triangle APR. We have cos RPA = tan RP. cot AP. Ience, denoting the hour angle APZ by t, the latitude by c, and the declination by 8, we have cos t = - tan q tan 8. (470) And the hour angle being known, the time may be found. In the case of the sun, the formula (470) gives the time from sunrise to noon, and hence the length of the day. 2~. Being given the declination and the latitude, to find the azimuth from the north at rising. Let A denote the required azimuth, then A = AR. Hence, from the triangle ARP, we have sin 8 = cos <j. cos A. (471) 3~. Being given the hour angle and declination of a star, tofind the azimuth and altitude. 152 Applications of Spherical Trigonometry. Let Z denote the zenith distance ZS, A the azimuth from the north, p the angle ZSP at the star; then, by Delambre's Analogies, cos Z. sin. (p +.) = cos 1t. cos i (8 - 4). (472) cos Z. cos ~ (p +- ) = sin- t. sin L (8 + ). (473) sin iZ. sin (p - A) = cos t. sin (8 - 4). (474) sin Z. cos (p - ) = sin t. cos 3 (8 + <). (475) Hence, when t, 8, 4 are given; that is, the hour angle and declination of a heavenly body, and the latitude of the observer, z, p, A can be found. In a similar manner may be solved the converse problem:-Given the azimuth and altitude, to find the hour angle and the declination. 4~. If a denote the altitude of the sun at 6 o'clock, and a' the altitude when east or west; then sin a = sin 8. sin <. (476) sin a'= sin 8 - sin 4. (477) EXERCISES. —XXXVI. 1. In latitude 45~ N., prove that the shadow at noon of a vertical object is three times as long when the sun's declination is 15~ S. as when it is 15~ N. 2. The altitude of a star when due east was 20~, and it rose EbN; required the latitude. 3. Given the sun's longitude, to find his right ascension and declination. S D 7 Fig. 61. Let a denote the right ascension, 8 the declination, o the obliquity of the ecliptic. Now let S denote the sun's place in the ecliptic rS. Draw Astronomy. 153 SD perpendicular to /TD, the equator; then, if x denote the longitude, we have the triangle r)SD, tan a = cos w tan A. (478) sin 5 = sin o sin A. (479) 4. Given the azimuth of the sun at setting, and at 6 o'clock, find the sun's declination, and the latitude. 5. If the sun's declination be 15~ N., and length of day four hours, prove tan 4 = sin 60~ tan 75~. sin,. tan À + sin 8. tan a 6. Prove that cos, = sin.tan +sin tan (480) sin q. tan a + sin. tan A 7. Given the sun's declination and the latitude, show how to find the time when he is due east. 8. If the sun rise N.E. in latitude 4, prove that cot hour angle at sunrise = - sin >. 9. Given the meridian altitude and altitude when east, find the latitude and the declination. 10. Given the right ascension and the declination of a star S, to find its latitude and longitude. Y D Fig. 62. Let )rD, rL be the equator and the ecliptic, S the star, SD, SL perpendicular to Tr, rL; then, if a be the right ascension, 8 the declination, I the 154 Applications of Spherical Trigonometry. latitude, and x the longitude of S, denoting the angle SrD by 0, from the right-angled triangles Sr-D,.SrL, we get tan A cos ( -c ) sin 1 sin(0 - Co) cot = sin a cot, = - --- (481) tan a cos 0 sinm sm O The first of these equations determines o, and the others x and 1. 11. Being given the latitudes and longitudes of two places on the earth considered as a perfect sphere, to find the distance between them. This is evidently a case of ~ 66, viz., when two sides and the contained angle are given, to find the third side. 12. Find the latitude, being given the declination, and the interval between the time the sun is west and sunset. 13. If the latitudes and longitudes of two places on the earth be given, show how to find the highest latitude attained by a ship in sailing along a great circle from one place to the other. 14. Being given the latitudes and longitudes of two places, find the sun's declination when he is on the horizon of both at the same instant. 15. If the difference between the lengths of the longest and the shortest day at a given place be six hours, find the latitude. 16. If two stars rise together at two places, prove that the places will have the same latitude; and if they rise together at one place, and set together at the other, the places will have equal latitudes of opposite names. 17. If pi, p2 be the radii vectors of two planets which revolve in circular orbits, prove, if when they appear stationary to one another, the cotangent of P2's elongation, seen from Pi, be 2 tan 0, that 2pi = p2 tan O. tan O. (482) 18. If 8 be the declination of a heavenly body, which in its diurnal motion passes in the minimum time from one to another of two parallels of altitude, whose zenith distances are Z, Z', prove that = cos (Z+ Z') sin ~ (c Z = Z ) -sin lat. (483) cos (Z - Z') v 19. If I be the latitude, w the obliquity of the ecliptic, prove that if the lengths of the shadow of an upright rod at noon on the longest and the shortest days be as 1: n, sin 21: sin 2w:: n + 1: n - 1. (484) Astronomy. 155 20. Determine the latitude, and the sun's declination, being given that the sun sets at 3 o'clock, and is 18~ below the horizon at 4 o'clock. 21. Determine the latitudes of two places A, B from the following data:When the sun is in the tropic of Cancer, he rises an hour earlier at A than at B; and when at the tropic of Capricorn, an hour earlier at B than at A. 22. If in latitudes 11, 12,3 on the same day, on the same meridian, the lengths of meridian shadows of towers of equal heights be s, sa, s3, prove S1 (S2 - 83)2 S2 (83 - S1)2 83 (S1 - S2)2 tan (12 - k) tan(l3 - I) tan (hl - 12) 23. If the time of the sun, being due east, be midway between sunrise and 12 o'clock, find the latitude, the declination being given. 24. If the sun be due east at a given place two hours after the rising, find the declination. 25. Given the right ascension and declination of four stars, find the right ascension and declination of the point in the heavens where the diagonals of the spherical quadrilateral which they determine intersect each other. 156 Miscellaneous Exercises. Miscellaneous Exercises. 1. Prove that in a right-angled spherical triangle tan r = sin (s - c), tan r'= sin (s-b), tan r" = sin (s-a), tan r"' = sin s. 2. If the plane angles of a trihedral angle be respectively equal to the angles of a square, a hexagon, and a decagon, prove that the sum of its dihedral angles is five right angles.-(CATALAN.) 3. If Ai be a chordal angle of a spherical triangle ABC, prove 1 + cos a - cos b - cos c cos -4A1 = --. (486) 4 sin b sin ( c 4. If a spherical quadrilateral be inscribed in a small circle of the sphere, prove that the cosine of its third diagonal is equal to the product of the cosines of the tangents drawn to the small circle from the extremities of the third diagonal. 5. Prove that the volume of the pyramid whose summits are the angular points of a spherical triangle and the centre of the sphere, if the radius be equal to unity, is l3/tan r. tan ra. tan rb. tan rc. 6. Prove that the angles of intersection of IHAT'S circle with the sides of a spherical triangle are (A - B), (B - C), (C - A), respectively. 7. If in a trihedral angle O-ABC we inscribe two spheres, which touch each other, if R1, R2 be their radii, prove that R2 4sin (s - a) sin(s - b) sin (s- c) 1 + sin (s - a) sin(s - b) sin(s -c) } (Ri sin s sins } (STEINER.) (487) 8. If any angle of a spherical triangle be equal to the corresponding angle of its polar triangle, prove sec2A + sec2B + sec2 C + 2sec A sec B sec C = 1. (488) 9. If ABC be a diametral triangle, of which the side c is the diameter, sin2 C = sin2 a + sin2 b. (489) 10. Express sin s, sin (s - a), &c., in terms of the in-radii of a triangle and its colunar triangles. 11. Express sinE, sin (A - E), in terms of the circumradii of a triangle and its colunars. Miscellaneous Exercises. 157 12. If x, tz denote the perpendiculars from the middle point of BC on the internal and external bisectors of the angle A, prove that 2 sin À sin j = n. sin ~ (B + C) sec a. (490) 13. If there be any system of fixed points Ai, 42, A3, &c., and a corresponding system of multiples 11, 12, 13, &c., and P a point satisfying the condition 2 (I cos AP) = constant, the locus of P is a circle. DEM.-Let x, y, z denote the normal co-6rdinates of P with respect to a fixed trirectangular triangle xi, yl, zl, &c., those of Ai, &c. Then (Art. 104) we have (Ixi). x + (ly) y + (lzI) z = constant. Put (Ixl) =X, (yl) =Y, (Izl) = Z; then, if O be a point whose normal co-ordinates are X Y Z y, R, 1, where 2 =X2+ y2+ 2), we have 2 ( cos AP) = R cos OP = constant. Hence the locus of P is a circle. Cor.-If: (i cos AP = 0, either OP =-, and the locus is a great circle, or. = 0, and then X, Y, Z must each separately vanish. 14. The sum of the cosines of the arcs, drawn from any point on the surface of a sphere to all the summits of an inscribed regular polygon, is equal to zero. 15. If O be the incentre of a spherical triangle ABC, prove that cos OA sin(b- c) +cos OB sin(c-a) + cos OC sin(a-b)= 0. (491) 16. If the side AB of a spherical triangle be given in position and magnitude, and the side AC in magnitude, prove, if BC meet the great circle, of which A is the pole in D, that the ratio cos BD: cos CD is constant. 17. The eight circles tangential to any three given circles on the sphere may be divided into two tetrads, say X, Y, Z, W; X', Y', Z', W', of which one is the inverse of the other, with respect to the circle, cutting the given circles orthogonally. 18. Any three circles of either tetrad, and the non-corresponding circle of the other tetrad, are touched by a fourth circle.-(HART.) 19. Any two circles of the first tetrad, and the two corresponding circles of the second, have a fourth common tangential circle.-(Ibid.) 158 Miscellaneous Exercises. *20. If A, B, C, D be four points on the same great circle, and if p be the angle of intersection of the small circles, whose spherical diameters are AC and BD, prove that the six anharmonic ratios of the points A, B, C, D are sin2 ~, cos2 ~, -tan2 ~, cosec2 (p, sec2-, -cot22. t21. The mutual power of two circles on the sphere is unaltered by inversion. 22. Prove the relation (412) by inversion. 23. If from a fixed point O on a great circle three pairs of arcs OA, OA'; OB, OB'; OC, OC' be measured, such that tan OA.tan OA'= tan OB. tan OB' -tan OC. tan OC' = k2, where k is a constant; then the anharmonic ratio of any four of the six points A, A', &c., which contains only one pair of conjugates, such as (ABCC'), is equal to the anharmonic ratio of their four conjugates (A'B'C'C).-(Compare Sequel to Euclid, p. 132.) Draw a tangent to the great circle, and produce the radii through the points A, A', &c., to meet the tangent. DEF. I.-A system of pairs of points, such as AA', BB', CC', fufi/ling the conditions that the anharmonic ratio of any four being equal to that of their four conjugates, is called a system in involution. DEF. II.-If two points D, D' be taken in opposite directions from O, such that tan2 OD = tan2 OD' = k2, each point being evidently its own conjugate, is called a double point. DEF. III.-If a system of points in involution on a great circle X be joined by arcs of great circles to any point P not on X, the six joining arcs having evidently the anharmonic ratio of the pencil formed by any four equal to that formed by their four conjugates, is called a pencil in involution. 24. The double points D, D' are anharmonic conjugates to any pair AA' of conjugate points. 25. The six arcs joining any point on a sphere to the intersection of the sides of a spherical quadrilateral form a pencil in involution. *This theorem in plano was first published by the author in the Philosophical Transactions, 1871, p. 704. t This theorem, in a different form, viz., "the ratio of the sine squared of half the common tangent of two small circles to the product of the tangents of their radii is unaltered by inversion," was first given by the author in a Memoir " On the Equations of Circles," in the Proceedings of the Royal Irish Academy, 1866. Miscellaneous Exercises. 159 26. Any great circle is cut in involution by the sides and diagonals of a spherical quadrilateral. 27. If two diagonals of a spherical quadrilateral be quadrants, the third is a quadrant. 28. Show that the method given in "Sequel," p. 121, for describing a circle touching three circles, may be extended to the sphere. 29. Inscribe in a spherical triangle or in a small circle a triangle whose sides shall pass through three given points. 30. Prove that if CC' be the symmedian drawn from the angle C of a spherical triangle, tan C' = 2 jcos2c - COs2 i (a + b) cos2 (a b) (492) cot a sin b + cot b sin a 31. If ABCD be a cyclic quadrilateral, and P any point in the circumcircle, prove that sinAPB. sin CPD sin A AB. sin ~ CD sinAPC. sin BPD sin 'AC. sin B) (4 32. If three great circles having two points common intersect the sides of a spherical triangle in angles ai, a2, a3; 81, $2, J3: 71, 72, 73, respectively, prove that COs al, COS a2, COS a3 cos B1i C, s 2, cs 3 = 0. (494) cs 71, COS 72, COS 73 33. Given the base of a spherical triangle and the two bisectors of the vertical angle, solve the triangle. 34. If two sides of a spherical triangle be given in position, and a point in the base fixed, if the base be bisected at the fixed point, prove that the area is either a maximum or a minimum. 35. If the sines of the perpendiculars let fall from a point on the sides of a spherical polygon, each multiplied by a given constant, be given, the locus of the point is a circle. 36. O, S are two points on the surface of the sphere; O is fixed, and S suffers a small displacement along OS proportional to sin OS; prove that the displacement estimated in the directions of two great circles at right angles to each other, passing through S, are proportional to the cosines of the distances of their poles from 0. 160 Miscellaneous Exercises. 37. If a chord PQ of a small circle whose spherical centre is C pass through a fixed point O on the sphere, prove that tan PCO. tan ~ OCQ is constant. 38. If through a given point O a great circle be drawn, cutting a small circle in the points A, A', and on it a point X, taken so that cot OX = cot OA + cot OA', the locus of X is a great circle, 39. If arcs which intersect in a point O be drawn from the angles of a triangle, meeting the opposite sides in the points dA', B', C', prove tan 0'O tan B'O tan C'O tan A'O + tan O + tan B'O + tan OB tan C'O + tan OC = (495 40. If a great circle, passing through a fixed point O, cut the sides of a spherical polygon in the points A, B, C, &c.; and if X be a point, such that cot OX = cot OA + cot OB + cot OC, &c., the locus of X is a great circle. INDEX. ALLMAN, Professor, 131. Altitudes of a spherical triangle, 23, 26. --- of a heavenly body, 150, 151, 152, 153, 154. Ambiguous case, 58, 59. Analogies between the geometry of the plane and sphere, 7. - Breitschnieder's, 47. -- Delambre's or Gauss's, 40, 41, 43, 47, 65. --- - Napier's, 39, 65. ----- Reidt's, 41, 59. Angle (the hour angle), 152. -— dihedral, 133, 134, 139, 141, 142. — horizontal, 143, 145. -- of intersection of two circles, 6, 18, 123, 125. - - solid, 10, 19, 127, 133, 138, 140. of a chordal triangle, 144. - - of a spherical triangle, 12. Anharmonic ratio, 68, 69, 71, 75, 110. Annales Nouvelles, theorems from, 43, 45, 90, 95, 136. Antipodal triangles, 10, 13, 93. Area of a lune, 15. - of a spherical polygon, 17, 97. -- of a spherical quadrilateral, 127. -- of a spherical triangle, 16, 93, 95, 147, 148. Aries, first point of, 151, 153. Ascension, right, 150, 152, 153, 155. Astronomy, definitions of, 149. ----- exercises on, 152. -- - problems on, 151. Axis of a circle, 4. -- of similitude, 104. Azimuth, 150, 151, 152, 153. Baltzer on Napier's ules, 36. Barbier, theorems by, 45. Breitschneider, theorems by, 47. Brianchon, theorem by, 111. Briot et Bouquet, method of using tables, 48. M 162 Index. Brùnnow, theorems by, 65. Cagnoli, theorem by, 86, 96. Cancer, tropic of, 155. Capricorn, tropic of, 155. Catalan, 156. Cauchy, method of solving spherical triangles, 63. Centre of circumcircle, 78. of excircle, 78. of incircle, 75. of inversion, 119. - of mean position, 81. of similitude, 103, 104. -- of sphere, 1. Circle, circumscribed, 78, 79, 80, 81. declination, 150. -- great and small, 2. --- Hart's, 83, 117. --- inscribed, 75, 76, 77, 80, 81, 83. -- Lexell's, 92, 93, 96, 97. -- vertical, 150. Chasles, theorem by, 121. Circular parts, Napier's, 35, 65. Circumradius of a tetrahedron, 135. Coaxal circles, 101, 106, 123. Colunar triangles, 10, 80, 85, 117, 124. Concurrent, 70, 72, 93, 94. Conjugates, isogonal, 73. - isotomic, 72. Crelle, Journal of Mathematies, 22, 47. Coordinates, normal, 73. - spherical, four systems of, in astronomy, 150. _ ----_ - triangular, 73. Cube, 129, 130, 131. Cyclic points, 127. Darboux, 113. Declination, 150-155. Delambre, 40, 41, 43, 65. De Morgan on Napier's rules, 36. Dodecahedron, 129, 131, 132. Dostor, solution by, 136. Diagonal of a parallelopiped, 134. Diametral triangle, 18, 81. Dupu;s, theorem by, 121. Ecliptic, 150, 152, 153, 154. Index. 163 Edge, 131, 132. Envelope, 83. Equator, 150, 152. Equinox, 150. Euler, 128. Excess (spherical), 13, 16, 30-32, 43, 44, 47, 55-57, 72, 75, 77, 78-81, 85-90, 95-98, 146-148, 156. Frobenius, theorems by, 113, 116. Gauss, 41. Geodesy, 143-148. Gergonne, 70. Geometry (spherical), 1-8. Girard, theorem by, 16, 17. Gudermann, theorem by, 90. Harmonic, 68, 71, 106-111. Hart, Sir Andrew Searle, theorems by, 83, 117, 156, 157. Homothetic, figures, 105. ----- points, 103, 104. Horizon, 149, 150. Hour angle, 150-152. Icosahedron, 129-131. Inclination, 129-131. Inversion, 105, 118-127. Involution, 158, 159. Keogh, theorem by, 95. Klein, 131. Lachlan on Frobenius's theorem, 114. Latitude, 150-155. Legendre, 144-149. Lemoine, 75. Lexell's circle, 92, 93, 95, 98. Lhuilier, 44. Lhuilierian, 44, 45, 56, 80. Limiting points, 101. Locus, 4, 9, 24, 26, 38 66, 82, 92, 102, 121, 160. Longitude, 150-155. Lozenge, 18. Lune, 10, 15, 97 Maximum area, 95, 159 Mean centre, 81-84. Meridian, 149-155. Mutual power of two circles on the sphere, 112-118. Nagel, 84. Napier, 35, 36, 39, 41, 65, 66. M 2 164 Index. Neuberg, theorems by, 22, 91, 94, 96, 97, 98, 103, 138, 139, 140. Norm, 22, 31. Oblique angle, reduction to the horizon, 143, 144. Obliquity of the ecliptic, 150-155. Octahedron, 129-131. Parallelogram, 18. Parallelopiped, 131-134. Parts of a triangle, 19. Pascal's theorem, 111. Points, double of involution, 158. homothetic, 103. inverse, 103, 123. Poles of a circle on the sphere, 3. - of the heavens, 149. - of the primitive, 122. Poles and polars, 106-111. Polyhedra, 128-142. Power, spherical, 100. mutual, 112-118. Primitive, 121-127. Proceedings of the Royal Irish Academy, 116, 158. Prouhet, proofs by, 43, 89. Quadrantal triangles, 81, 112. Ratio of section, 67. - anharmonic, 68, 69, 70, 158. Radical circle, 101, 102. Roy, application of Girard's theorem, 117. Rule, mnemonic, Napier's, 35. - of transformation, 32, 85. Roy's, 148. Salmon, theorem by generalized, 111. Serret, applications of stereographie projection, 124-127. Similitude, axes of, 104. centres of, 103-106. Solution, oblique-angled triangles, 54-64. ---- - right-angled triangles, 49-54. Sphere, elementary properties, 1-8. Staudt, definition of sine of solid angle, 22. Staudtians, 22, 23, 132, 140. Steiner, theorems, 72, 75, 81, 92, 93, 95, 127, 156. Stereographic projection, 121-128. Stewart, theorem by, generalised, 24. Supplemental triangles, 11. Symmedian lines of a triangle, 159. Index. 165 Symmedian point of a triangle, 75. Tetrahedra, 128-142. Transactions, Philosophical, 158. Transversal isogonal, 73. -__ —_- isotomic, 73. Triangle, chordal, 144. - spherical, 9. Volume of a parallelopiped, 131. ----- of a tetrahedron, 132, 141. Vertical circle, prime, 150. Wallis, 17. Wolstenholme, 141. Zenith, 150, 152, 154. THE END. Recently Published, Crown 8vo, 276 pp., Cloth, 7/6. A TREATISE ON PLANE TRIGONOMETRY, CONTAINING AN ACCOUNT 0F HYPERBOLIC FUNCTIONS, WITH BY JOHN CASEY, LL.D., F.R.S., Fellow of the Royal University of Ireland; Member of the Council of the Royal Irish Academy, &c., &c. Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co. EXTRACTS FROM CRITICAL NOTICES. From the " LYCEUM," May, 1888.. "There is a great deal of new matter not hitherto introduced into text-books on trigonometry, at least in these countries -matter which is not only interesting in itself, but important on account of the present requirements of the higher mathematical teaching.... The author has given many proofs of well-known theorems, as in the case of the four fundamental formulae. Some of these are original, others are contributed by distinguished correspondents, and last, though not least, many have been derived by very extensive reading from the writings of wellknown authors, to whom reference is always carefully made. Indeed this last fact adds a great interest from an historical point of view to all Dr. Casey's works. It is a decided mistake, as well as an injustice, to ignore the merits of other writers, and to appropriate their work without due acknowledgment, as is too often done. Many theorems owe much of their interest to the fact that they are associated with the names of Euler, Jacobi, Lagrange, Cauchy, or even stars of a much lesser magnitude...." I From "NATURE," July 5, 1888. "Dr. Casey's ' Treatise on Plane Trigonometry' is quite independent of the ' Elementary Trigonometry' by the same author. It is a most comprehensive work, and quite as exhaustive as any ordinary student will require. Dr. Casey shows his usual mastery of detail, due to thorough acquaintance, from long teaching, with all the cruces of the subject. He has embraced in his pages all the usual topics, and has introduced several points of extreme interest from the best foreign text-books. A very rigid proof is given of the exponential theorem, and a section is devoted to interpolation.... Chapters v. and vi., which are devoted to triangles and quadrilaterals, are exceedingly interesting, and contain quite a crop of elegant propositions culled from many fields. Following the course adopted by other recent writers, he gives a systematic account of imaginary angles and hyperbolic functions. 'The latter are very interesting, and their great and increasing importance, not only in pure mathematics but in mathematical physics, makes it essential that the student should become acquainted with them.' We may remark that Dr. Casey adopts the following notation: sh, ch, th, coth, sech, cosech, for sin h, cos h, &c., and has gone further than his English predecessors in introducing at this early stage the angle T, Hoùel's hyperbolic amplitude of 0 (r = amh. 0). Numerous illustrative examples and tables afford practice to the student in this branch... The special results, which on Dr. Casey's useful plan are numbered consecutively, reach 810. The book is rich in examples, and will be sure to find for itself a place on the mathematician's shelves, within easy reach of his hand." FrIom the "ATHENETUM," July 21, 1888. "Dr. Casey is no mere compiler. His heart is evidently in his work, and nearly every page of it bears the stamp of his individuality. The space at our disposal does not allow us to enter into details, but we can conscientiously say that we know of no work on plane trigonometry which contains so much new and useful matter, or which contains old matter better treated... The most interesting chapter is the last, which gives an exposition of imaginary angles and of hyperbolic functionsnovelties, we believe, hitherto in trigonometrical text-books, though not in mathematical periodicals. The hyperbolic functions are not only interesting from their close resemblance to the ordinary circular functions, but also important from their increasing utility in physical problems-two good and sufficient reasons for placing them early before mathematical students.... PFrom the "ACADEMY," Sept. 22, 1888. "Dr. Casey's object has been to write a work which shall be abreast of the best text-books on the subject, and in this he has succeeded. No difficulties are slurred over; in fact, the demonstrations are full, accurate, and complete. The text is amply illustrated by a rich collection of exercises. Not only have preceding text-books been consulted, but considerable contributions have been levied upon memoirs in mathematical journals, 2 and collections of problems (such as Wolstenholme's). Chapters v. and vi. (on triangles and quadrilaterals) contain an exceedingly interesting store of results, numbered for reference in the manner the writer has adopted in his previous books.... Adopting a practice introduced in one or two recent works on the subject, Dr. Casey assigns a sufficient space to the explanation of the hyperbolic series and cosines, and introduces some other functions to the student. It will be inferred that the present work is independent of the author's small introductory bookin fact, no reference whatever, we believe, is made to it. This treatise contains everything that one could expect, and, besides, has fresh matter-a section on interpolation, and one or two other small things-which we have not hitherto come across in similar works." From the " EDUCATIONAL TIMES," OCt. 1, 1888. "This treatise is very comprehensive, and quite sustains the author's reputation as a writer of mathematical text-books. While including all the usual propositions, Dr. Casey has, as usual, found room for much interesting matter derived from continental writers. The exercises are made to introduce much of the modern geometry of the triangle, and the chapters on triangles and quadrilaterals, one of the main features of the work, contain a large number of elegant and useful propositions. Imaginary angles and hyperbolic functions are fully treated, while an innovation is made by introducing the angle r, Hoiel's hyperbolic amplitude of 0. Dr. Casey fully recognises the value of these functions, 'and their great and increasing importance not only in pure mathematics but in mathematical physics.'.. From the " PRACTICAL TEACHER," Oct. 1888. "The book which we have before us contains, we believe, the most remarkable and complete treatment of its subject which has yet appeared in the English language. Too many writers have supposed that a knowledge of trigonometry only was necessary to enable them to write a book thereupon, and it has been rare, indeed, that a writer in every direction as competent as Dr. Casey, or with a mathematical eyesight so far-reaching, has grappled with an elementary subject like the present. Dr. Casey, fortunately for us, was known as an eminent mathematician before he became a writer of text-books. His investigations into geometry and higher algebra have gained him a European reputation; and when the first of his class text-books, the ' Sequel to Euclid,' appeared some years ago, we welcomed it in these columns as 'extraordinarily neat,' and extremely satisfactory even in the form it then assumed, which has since, in subsequent editions, been improved almost out of all likeness to its former self. This was followed in the course of last year by an equally remarkable treatise on analytical geometry, and it is little to be wondered at, therefore, if we open the present volume with very high anticipations. Nor are we disappointed. Alike in grasp and clearness, this book outdistances its only real rival, the venerable Todhunter. Of course we cannot help differing in a 3 few minor points from our author; but, on the whole, we are more at one with him than we have been with any previous author. " The examples-more than a thousand in number-are remarkably well-chosen and accurate. They are adapted not only to form a Trigonometrician, but also to develope a comprehensive mathematical talent. Many of them are original, and at least one-third have not appeared in any previous text-book. " In every case, some are solved as specimens (a point in which all Todhunter's works, by the way, are very deficient), and the answers of the remainder are given at the end of the book, where we have the welcome novelty of a tolerably complete index of the subject-matter, and the authors' names.... We repeat that it is not only an eminently good book, but that it is the best book on the subject." From "MATHESIS," Oct., 1888. "Ainsi qu'il le dit dans la Préface, M. Casey s'est proposé de composer un Traité de Trigonométrie qui soit en harmonie avec les récents manuels les plus avancés sur les autres branches des mathématiques: démonstrations rigourenses et complètes, ordre méthodique, développement clair et suffisant des connaissances faisant habituellement partie des traités de trigonométrie, indication des questions les plus initéressantes que l'on rencontre dans les journaux periodiques.... Les ouvrages des maîtres français (Serret, Briot et Bouquet, &c.), se distinguent par la recherche de la rigueur scientifique; les manuels anglais se recommandent par la richesse des matières traiteés ou proposées en exercises, et par un texte concis qui introduit rapidement le lecteur dans les régions élevées de la science, en glissant quelque peu sur les difficultées des éléments. Lé savant professeur de Dublin, parfaitement au courant de la littérature mathématique de l'Angleterre et de la France, a cherché à réunir dans une juste mesure, les qualités qu'on rencontre chez les auteurs renommés de l'un ou l'autre pays. I1 a pleinement réussi dans cette ~uvre difficile... La partie qui se rapporte au programme des établissements belges et français du degré moyen, est admirablement développée... Aucun détail de quelque importance n'est omis; ou trouve aussi bien les notions indispensables pour connaître let but proprement dit de la trigonométrie, que les ressources fournies par cette science aux autres branches des mathématiques... Comme nous l'avons déjà dit (Mathesis, t. viii., p. 114), ce traite de M. Casey constitue un repertoire très complet de trigonométrie: ou y retrouve, soit dans le texte, soit dans les exercises, toutes les questions intéressantes, publiées depuis un demi-siècle dans les journaux mathématiques. Dans les trente dernières années, l'enseignement de la trigonométrie plane a fait, en France des progrès considérables; ou en peut juger par les problèmes posés aux examens publics (St. Cyr, baccalauréat, école polytechnique, &c.), problèmes que certains journaux (par example le J. M. E. de M. de Longchamps) et même des recuiels spéciaux (publiés par les librairies iNony ou Croville-Morant) reproduisent régulièrement; M. Casey a en la bonne idée de faire un bon choix parmi ces problèmes pour ses exercices. En même temps, il a proposé un tres grand nombre de questiones originales." 4 Recently published, Crown 8vo, 350 pp., cloth, 7/6. A TREATISE ON THE ANALYTICAL GEO0METRY OF THE POINT, LINE, CIRCLE, AND CONIC SECTIONS, CONTAINING ûn <omut itaof most tent ^mmwsf WI IT NUMEROUS EXAMPLEES. BY JOIHN CASEY, LL.D., F.R.S., Fellow of the Royal University of lreland; Member of the Council of the Royal Irish Academy; &c., &c. Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co. OPINIONS OF THE WORK. From GEORGE GABRIEL STOKES, President of the loyal Society. " I write to thank you for your kindness in sending me your book on the 'Analytical Geometry of the Point, Line, Circle, and the Conic Sections.' " I have as yet only dipped into it, being for the moment very much occupied. Of course from the nature of the book there is much that is elementary in it; still I see there is much which I should do well to study." From PROFESSOR CAYLEY, Cambridge. "I have to thank you very much for the copy you kindly sent me of your treatise on 'Analytical Geormetry.' I am glad to see united together so many of your elegant investigations in this subject." From M. H. BROcARD, Capetainle du Genie à Montpellier. "J'ai en hier l'agréable surprise sur laquelle je me complais d'ailleurs depuis un certain temps de recevoir votre nouveau 'Traite de Geométrie Analytique.' "Je me suis emerveillé du soin que vous avez apporté à la redaction de cet ouvrage, et je forme dès aujourd'hui les voeux les plus sincères pour que de nombreuses editions de ce livre se répandent rapidement dans le public mathematique et parmi la jeunesse studieuse de nos universités." 5 From EDWARD J. ROUTH, M.A., F.R.S., &c., Cambridge. " It seems to me that so excellent a treatise will soon make its way to the front, even against the severe competition which all books on Conics now meet with. I fancy that many of the theorems cannot be elsewhere found so conveniently explained." From the " FREEMAN'S JOURNAL." "This treatise exhibits in a marked degree the qualities which distinguish the author's other works. It is at once compact and comprehensive... Dr. Casey's treatise, indeed, may well accomplish for this generation what Dr. Salmon's did for their fathers, namely, to introduce the young mathematician to the latest developments in the highest departments of the Science.... Scarcely any important step is there in the work which he has not simplified, giving one, and sometimes several, original methods.... The method of projection is treated of in an entirely original manner, not requiring the consideration of space of three dimensions, the projection being performed in one plane, and the equation of the curve obtained by a simple transformation. The properties of the M'Cay, Neuberg, and Brocard Circles are considered at some length: indeed the latter has now a literature of its own, to which Dr. Casey has largely contributed. There are considerably over a thousand examples, graduated from easy applications of the formulæe to exercises of the highest class of difficulty. Many of these are original, and many are historically interesting." From the " DUBLIN EVENING MAIL." "This work will give a new impulse to the study of Analytic Geometry, introduce the student to new and more powerful methods, and greatly enlarge his mathematical horizon. It has been known for some time that Dr. Casey was engaged on a 'Conic Sections,' and people expected that notwithstanding the many works in the field, Dr. Casey's would present a good many valuable novelties; but the work has, we venture to think, excelled all anticipation.... Dr. Casey makes a large use of determinants. He introduces them in the very first chapter, and employs them to the end. In the first chapter we find a section devoted to complex variables, and Gauss's geometrical representation of them; and from Clebsch he takes the comparison of ' point' and ' line' co-ordinates. He gives a great development to trilinear co-ordinates and tangential equations, discusses the new circles (Tucker's, Brocard's, Neuberg's, M'Cay's), applies Aronhold's notation to the discussion of the general equation of the second degree, and reproduces from his own Papers in the Transactions of learned societies many important theorems and problems relating to inscribed and circumscribed figures, orthogonal conics, and the tact-invariant of two-conics.... The work is a noble monument to Dr. Casey's genius, and his mastering of all the resources of Modern Analytic Geometry." 6 From "NATURE." "Dr. Casey, by the publication of this third treatise, has quite fulfilled the expectations we had formed when we stated some months since that he was engaged upon its compilation. It is a worthy companion of those which have preceded it. It possesses many points of novelty, i. e. for the English mathematician. He has from the first introduction of certain recent continental discoveries in geometry taken a warm interest in them, and in the purely geometrical treatment of them, has himself given several beautiful proofs, and has added discoveries of his own. We may here note that this work has met with a very warm welcome in France and Belgium. The author himself has added so much in years now long past to several branches of the subject treated of in the volume under notice-the equation of the circle (and of the conic) touching three circles (three conics), and other properties-that he is specially fitted, by his intimate acquaintance with it, and by his long tuitional experience, to write a book on analytical geometry." From the " EDUCATIONAL TIMES," September, 1886. "In this book the author has added to those propositions usually met with in Treatises on Analytical Geometry many which we have seen in no other books on the subject; notably extending the equations of circles inscribed in and circumscribed about triangles to polygons of any number of sides, and extending to Conics the properties of circles cutting orthogonally. The demonstrations are concise and neat. In many cases the author has substituted original methods of proof advantageously, and in some has also added the old methods. We would specially note his treatment of the General Equation of the Second Degree, which is more satisfactory than many we have seen. Throughout the book there are numerous exercises on the subject matter, and at the end of each section a collection of problems bearing on that part of the subject. These problems have been obtained from Examination Papers and other sources. They have been selected with much care and judgment. The name of the proposer has in many cases been added, and will cause more interest to be taken in the solution." 7 Seventh Edition, Price 4/6; or in two parts, each 2/6. THE FIRST SIX BOOKS OF THE ELEMENTS OF EUCLID, With Copious Annotations and Numerous Exercises. BY JOHN CASEY, LL.D., F.R.S., Fellow of the Royal University of Ireland; Member of the Council of the Royal Irish Academy; &o., &c. Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co. OPINIONS OF THE WORK. The following are a few of the Opinions received by Dr. Casey on this Work:From the REV. R. TOWNSEND, F.T.C.D., &c. "I have no doubt whatever of the general adoption of your work through all the schools of Ireland immediately, and of England also before very long." From the " PRACTICAL TEACHER." "The preface states that this book 'is intended to supply a want much felt by Teachers at the present day-the production of a work which, while giving the unrivalled original in all its integrity, would also contain the modern conceptions and developments of the portion of Geometry over which the elements extend.' " The book is all, and more than all, it professes to be.... The propositions suggested are such as will be found to have most important applications, and the methods of proof are both simple and elegant. We know no book which, within so moderate a compass, puts the student in possession of such valuable results. " The exercises left for solution are such as will repay patient study, and those whose solution are given in the book itself will suggest the methods by which the others are to be demonstrated. We recommend everyone who wants good exercises in Geometry to get the book, and study it for themselves." 8 From the "l EDUCATIONAL TIMES." "The editor has been very happy in some of the changes he has made. The combination of the general and particular enunciations of each proposition into one is good; and the shortening of the proofs, by omitting the repetitions, so common in Euclid, is another improvement. The use of the contra-positive of a proved theorem is introduced with advantage, in place of the reductio ad absurdum; while the alternative (or, in some cases, substituted) proofs are numerous, many of them being not only elegant but eminently suggestive. The notes at the end of the book are of great interest, and much of the matter is not easily accessible. The collection of exercises, 'of which there are nearly eight hundred,' is another feature which will commend the book to teachers. To sum up, we think that this work ought to be read by every teacher of Geometry; and we make bold to say that no one can study it without gaining valuable information, and still more valuable suggestions." Prom the "JOURNAL OF EDUCATION," Sept. 1, 1883. ( In the text of the propositions, the author has adhered, in all but a few instances, to the substance of Euclid's demonstrations, without, however, giving way to a slavish following of his occasional verbiage and redundance. The use of letters in brackets in the enunciations eludes the necessity of giving a second or particular enunciation, and can do no harm. Hints of other proofs are often given in small type at the end of a proposition, and, where necessary, short explanations. The definitions are also carefully annotated. The theory of proportion, Book V., is given in an algebraical form. This book has always appeared to us an exquisitely subtle example of Greek mathematical logic, but the subject can be made infinitely simpler and shorter by a little algebra, and naturally the more difficult method has yielded place to the less. It is not studied in schools; it is not asked for even in the Cambridge Tripos; a few years ago, it still survived in one of the College Examinations at St. John's; but whether the reforming spirit which is dominant there has left it, we do not know. The book contains a very large body of riders and independent geometrical problems. The simpler of these are given in immediate connexion with the propositions to which they naturally attach; the more difficult are given in collections at the end of each book. Some of these are solved in the book, and these include many well-known theorems, properties of orthocentre, of nine-point circle, &c. In every way this edition of Euclid is deserving of commendation. We would also express a hope that everyone who uses this book will afterwards read the same author's ' Sequel to Euclid,' where he will find an excellent account of more modern Geometry." Second Edition, Price 6s. A KEY TO THE EXERCISES IN THE FIRST SIX BOOKS OF CASEY'S "ELEMENTS OF EUCLID." 9 BY THE SAME AUTHOR. Fifth Edition, Revised and Enlarged. 3s. 6d., Cloth. A SEQUEL TO THE FIRST SIX BOOKS OF THE ELEMENTS OF EUCLID. Dublin: Hodges, Figgis, & Co.1 London: Longmans, Green, & Co. EXTRACTS FROM CRITICAL NOTICES. From the " SCHOOL GUARDIAN." "This book is a well-devised and useful work. It consists of propositions supplementary to those of the first six books of Euclid, and a series of carefully arranged exercises which follow each section. More than half the book is devoted to the Sixth Book of Euclid, the chapters on the ' Theory of Inversion' and on the' Poles and Polars' being especially good. Its method skilfully combines the methods of old and modern Geometry; and a student, well acquainted with its subject-matter, would be fairly equipped with the geometrical knowledge he would require for the study of any branch of physical science." From the " PRACTICAL TEACHER." "Professor Casey's aim has been to collect within reasonable compass all those propositions of Modern Geometry to which reference is often made, but which are as yet embodied nowhere.... We can unreservedly give the highest praise to the matter of the book. In most cases the proofs are extraordinarily ncat... The notes to the Sixth Book are the most satisfactory. Feuerbach's Theorem (the nine-points circle touches inscribed and escribed circles) is favoured with two or three proofs, all of which are elegant. Dr. Hart's extension of it is extremely well proved.... We shall have given sufficient commendation to the book when we say that the proofs of these (Malfatti's Problem, and Miquel's Theorem), and equally complex problems, which we used to shudder to attack, even by the powerful weapons of analysis, are easily and triumphantly accomplished by Pure Geometry. " After showing what great results this book has accomplished in the minimum of space, it is almost superfluous to say more. Our author is almost alone in the field, and for the present need scarcely fear rivals." From the " JOURNAL OF EDUCATION." "Dr. Casey's ' Sequel to Euclid' will be found a most valuable work to any student who has thoroughly mastered Euclid, and imbibed a real taste for geometrical reasoning... The higher methods of pure geometrical demonstration, which form by far the larger and more important portion, are admirable; the propositions are for the most part extremely well given, and will amply repay a careful perusal to advanced students." IO From "l MATHESIS," April, 1885. "A Sequel to Euclid de M. J. Casey est un de ces livres classiques dont le succès n'est plus à faire. La première édition a paru en 1881, la seconde en 1882, la troisième en 1884, et l'on peut prédire sans crainte de se tromper, qu'elle sera suivie de beaucoup d'autres. C'est un ouvrage analogue aux Théorèmes et Problèmes de Géométrie de M. Catalan, et il a les mêmes qualités: il est clair, concis, et renferme beaucoup de matières, sous un petit volume... Le sixième livre de louvrage de Casey est aussi étendu à lui seul que les quatre premiers. Il occupe les pages 67 à 158, c'est-à-dire la moitié du volume. C'est en réalité une Introduction à la Géorétrie Supérieure. Dans cette partie de l'ouvrage, on rencontre des demonstrations d'une rare élégance dues à M. Casey lui-même." From " MATHtESIS," December, 1886. "La troisième édition, publiee en 1884, et dont il a été rendu compte dans ce journal (t. v. 1885, p. 76-78) fait aujourd'hui place à une quatrième, dont la première partie, renfermant 164 pages, 152 figures et 293 exercices, est la reproduction de six chapitres de la précédente édition, car l'auteur n'a pas eu le temps d'en remanier la substance; mais toute la seconde partie, 58 pages, 23 figures et 181 exercices, intitulée: Géométrie récente du triangle, est absolument nouvelle et complétement refondue. "Elle se divise en six sections consacrées aux études suivantes: 1~. Théorie des points isogonaux et isotomiques, des lignes antiparallèles et des symédianes; 2~. théorie des figures directement semblables; 3~. cercles de Lemoine et de Tucker; 4~. théorie générale d'un système de trois figures semblables; 5~. applications particulières de la théorie des figures directement semblables (au cercle de Brocard et au cercle des neuf points) 6~. théorie des polygones harmoniques. " Chacune de ces subdivisions, notamment la dernière, est suivei d'exercices intéressants, théorèmes à démontrer ou problèmes à résoudre, parmi lesquels nous devons signaler à l'attention des géomètres les élégantes généralisations dues à MM. Casey, G. Tarry et J. Neuberg dans l'étude des polygones harmoniques. " Nos lecteurs sont déjà familiarisés avec ces nouvelles théories, et ils connaissent aussi l'éminent géomètre qui, depuis trente ans, fait autorité dans l'enseignement des universités de la GrandeBretagne. L'ouvrage est à la hauteur de la réputatio n du maître, et à tous les degrés d'avancement de leurs études, les élèves y trouveront un guide précieux et instructit. Mais cet ouvrage n'est pas seulement utile aux étudiants auxquels il s'addresse; les professeurs y rencontreront des démonstrations nouvelles, une abondante variété de problèmes (près de cinq cents (réellement attrayants et judicieusement choisis, enfin une exposition aussi claire que possible des derniers, progrès de la géométrie du triangle, présentés avec toutes les simplifications désirables. " Les souhaits que nous avions exprimés au sujet de la troisiéme édition se sont trouvés bientôt réalisés; nous ne pouvons mieux faire que de les reproduire aujourd'hui, persuadé que le nouvel ouvrage sera, comme il le mérité, chaleureusement accueilli et justement apprécié par le public mathématique." II Second Edition, Price 3s., post free. A TREATISE ON ELEMENTARY TRIGONOMETRY. BY JOHN CASEY, LL.D., F.R.S., Fellow of fte Royal University of Ireland; Corresponding Member of the Royal Society of Liege; &c., &c. Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co. OPINIONS OF THE WORK. From SIR ROBERT S. BALL, Astronomer Royal of Ireland. " So far as I have examined it, it appears to be a gem; and, like all your works, it has the stamp on it." From W. J. KNIGHT, LL.D., Cork. "Your work on Elementary Trigonometry is, in simplicity and elegance, superior to any I have seen. I shall introduce it immediately into my classes." From the "LONDON INTERMEDIATE ARTS GUIDE." "This is just the kind of book a student requires, as it represents almost exactly the amount of reading expected of a candidate for the Intermediate Arts, graduated gently, and with a copious supply of examples." From the " EDUCATIONAL TIMES." "This is a Manual of the parts of Trigonometry which precede De Moivre's theorem. All necessary explanations are given very fully, and the Exercises are very numerous, and are carefully graduated." Price 3s. KEY TO THE EXERCISES IN THE "TREATISE ON ELEMENTARY TRIGONOMETRY." LONDON: LONGMANS & CO. DUBLIN: HODGES, FIGGIS, & CO. 12