LIBKAKY OF THE UNIVERSITY OF CALIFORNIA. PHYSICS DEPARTMENT. (HIT (')!' Miss ROSE WHITING. ''fycetieil September, 1896. No. SPHERICAL TRIGONOMETRY. SPHERICAL TRIGONOMETRY Jfor % to of Colleges an& jfejpwrk WITH NUMEROUS EXAMPLES. BY I. TODHUNTER, M.A., F.E.S., HONOBAEY FELLOW OP ST JOHN'S COLLEGE, CAMBRIDGE. FOURTH EDITION. VIX7JBSIT7 MACMILLAN AND CO. 1878 [All Eights reserved.] (rfp / Camortoge : PRINTED BY C. J. CLAY, M.A. AT THE UNIVERSITY PRESS. PREFACE. work is constructed on the same plan as my treatise on Plane Trigonometry, to which it is intended as a sequel; it contains all the propositions usually included under the head of Spherical Trigonometry, together with a large collection of examples for exercise. In the course of the work reference is made to preceding writers from whom assistance has been obtained ; besides these writers I have consulted the treatises on Trigonometry by Lardner, Lefebure de Fourcy, and Snowball, and the treatise on Geometry published in the Library of Useful Knowledge. The examples have been chiefly selected from the University and College Examination Papers. In the account of Napier's Rules of Circular Parts an explanation has been given of a method of proof devised by Napier, which seems to have been overlooked by most modern writers on the subject. I have had the advantage of access to an unprinted Memoir on this point by the late R. L. Ellis of Trinity College ; Mr Ellis had in fact rediscovered for himself Napier's own method. For the use of this Memoir and for some valuable references on the subject I am indebted to the Dean of Ely. Considerable labour has been bestowed on the text in order to render it comprehensive and accurate, and the exam- ples have all been carefully verified ; and thus I venture to hope that the work will be found useful by Students and Teachers. L TODHUNTER. ST JOHN'S COLLEGE, August 15, 1859. IN the third edition I have made some additions which I hope will be found valuable. I have considerably enlarged the discussion on the connexion of Formula in Plane and Spherical Trigonometry ; so as to include an account of the properties in Spherical Trigonometry which are analogous to those of the Nine Points Circle in Plane Geometry. The mode of investigation is more elementary than those hitherto employed; and perhaps some of the results are new. The fourteenth Chapter is almost entirely original, and may de- serve attention from the nature of the propositions them- selves and of the demonstrations which are given. CAMBRIDGE, July, 1871. CONTENTS. CHAP. PAGB I. Great and Small Circles 1 IL Spherical Triangles ......... 6 III. Spherical Geometry 10 IV. Eelations between the Trigonometrical Functions of the Sides and the Angles of a Spherical Triangle 16 V. Solution of Eight-angled Triangles 32 VI. Solution of Oblique-angled Triangles 46 VH. Circumscribed and Inscribed Circles 59 Vin. Area of a Spherical Triangle. Spherical Excess . . . .67 IX. On certain approximate Formulas 75 X. Geodetical operations 85 XI. On small variations in the parts of a Spherical Triangle . . 93 XH. On the connexion of Formulae in Plane and Spherical Trigo- nometry 96 Xin. Polyhedrons 114 XIV. Arcs drawn to fixed points on the Surface of a Sphere . . 125 XV. Miscellaneous Propositions 135 XVI. Numerical Solution of Spherical Triangles .... 149 Of UITI7BRSIT7 U1X7JRSITY I. GREAT AND SMALL CIRCLES. 1. A SPHERE is a solid bounded by a surface every point of which is equally distant from a fixed point which is called the centre of the sphere. The straight line which joins any point of the surface with the centre is called a radius. A straight line drawn through the centre and terminated both ways by the surface is called a diameter. 2. The section of the surface of a sphere made by any plane is a circle. Let AB be the section of the surface of a sphere made by any plane, the centre of the sphere. Draw 00 perpendicular to the plane ; take any point D in the section and join OD } CD. Since 00 is perpendicular to the plane, the angle 00 D is a right angle ; therefore OD=J(OD 2 - OC 2 ). Now and are fixed points, so that 00 is constant j and OD is constant, being the radius of the T. S. T. B 2 GREAT AND SMALL CIRCLES. sphere ; hence CD is constant. Thus all points in the plane section are equally distant from the fixed point C ; therefore the section is a circle of which C is the centre. 3. The section of the surface of a sphere by a plane is called a great circle if the plane passes through the centre of the sphere, and a small circle if the plane does not pass through the centre of the sphere. Thus the radius of a great circle is equal to the radius of the sphere. 4. Through the centre of a sphere and any two points on the surface a plane can be drawn ; and only one plane can be drawn, except when the two points are the extremities of a diameter of the sphere, and then an infinite number of suet planes can be drawn. Hence only one great circle can be drawn through two given points on the surface of a sphere, except when the points are the extremities of a diameter of the sphere. When only one great circle can be drawn through two given points, the great circle is unequally divided at the two points ; we shall for brevity speak of the shorter of the two arcs as the arc of a great circle joining the two points. 5. The axis of any circle of a sphere is that diameter- of the sphere which is perpendicular to the plane of the circle ; the ex- tremities of the axis are called the poles of . the circle. The poles of a great circle are equally distant from the plane of the circle. The poles of a small circle are not equally distant from the plane of the circle ; they may be called respectively the nearer and fur- ther pole ; sometimes the nearer pole is for brevity called the pole. 6. A pole of a circle is equally distant from every point of the circumference of the circle. Let be the centre of the sphere, AB any circle of the sphere, C the centre of the circle, P and P' the poles of the circle. Take any point D in the circumference of the circle ; join CD, OD, PD. Then PD = J(PC 2 +CD 2 ) ; and PC and CD are constant, therefore PD is constant. Suppose a great circle to pass through the points P and D ; then the chord PD is constant, and therefore the arc of GREAT AND SMALL CIRCLES. 3 a great circle intercepted between P and D is constant for all positions of D on the circle AB. Thus the distance of a pole of a circle from every point of the circumference of the circle is constant, whether that distance be measured by the straight line joining the points, or by the arc of a great circle intercepted between the points. 7. The arc of a great circle which is drawn from a pole of a great circle to any point in its circumference is a quadrant. Let P be a pole of the great circle ABC ; then the arc PA is a quadrant. For let be the centre of the sphere, and draw PO. Then PO is at right angles to the plane ABC, because P is the pole of ABC, therefore POA is a right angle, and the arc PA is a quad- rant. B2 4 GREAT AND SMALL CIRCLES. 8. The angle subtended at the centre of a sphere by the arc of a great circle which joins the poles of two great circles is equal to the inclination of the planes of the great circles. B Let be the centre of the sphere, CD, CE the great circles in- tersecting at (7, A and B the poles of CD and CE respectively. Draw a great circle through A and B, meeting CD and CE at M and N respectively. Then AO is perpendicular to OC, which is a straight line in the plane OCD ; and BO is perpendicular to OC, which is a straight line in the plane OCE ; therefore OC is perpen- dicular to the plane AOB (Euclid, xi. 4); and therefore OC is perpendicular to the straight lines OM and ON, which are in the plane AOB. Hence MON is the angle of inclination of the planes OCD and OCE. And the angle AOB = AOM-BOM=BON-BOM= MON. 9. By the angle between two great circles is meant the angle of inclination of the planes of the circles. Thus, in the figure of the preceding Article, the angle between the great circles CD and CE is the angle MON. In the figure to Art. 6, since PO is perpendicular to the plane ACB, every plane which contains PO is at right angles to the plane ACB. Hence the angle between the plane of any circle and the plane of a great circle which passes through its poles is a right angle. GREAT AND SMALL CIRCLES. 5 10. Two great circles bisect each other. For since the plane of each great circle passes through the centre of the sphere, the line of intersection of these planes is a diameter of the sphere, and therefore also a diameter of each great circle ; therefore the great circles are bisected at the points where they meet. 11. If the arcs of great circles joining a point P on the surface of a sphere with two other points A and C on the surface of the sphere, which are not at opposite extremities of a diameter, be each of them equal to a quadrant, P is a pole of the great circle through A and C. (See the figure of Art. 7.) For suppose PA and PC to be quadrants, and the centre of the sphere ; then since PA and PC are quadrants, the angles POC and POA are right angles. Hence PO is at right angles to the plane AGO, and P is a pole of the great circle AC. 12. Great circles which pass through the poles of a great circle are called secondaries to that circle. Thus, in the figure of Art. 8 the point C is a pole of ABMN, and therefore CM and ON are parts of secondaries to ABMN. And the angle between CM and CN is measured by MN ; that is, the angle between any two great circles is measured by the arc they intercept on the great circle to which they are secondaries. 13. If from a point on the surface of a sphere there can be drawn two arcs of great circles, not parts of the same great circle, the planes of which are at right angles to the plane of a given circle, that point is a pole of the given circle. For, since the planes of these arcs are at right angles to the plane of the given circle, the line in which they intersect is per- pendicular to the plane of the given circle, and is therefore the axis of the given circle ; hence the point from which the arcs are drawn is a pole of the circle. 6 SPHERICAL TRIANGLES. 14. To compare the arc of a small circle subtending any angle at the centre of the circle with the arc of a great circle subtending the same angle at its centre. Let db be the arc of a small circle, G the centre of the circle, P the pole of the circle, the centre of the sphere. Through P draw the great circles PaA and PbB, meeting the great circle of which P is a pole, at A and B respectively ; draw Ca, Cb, OA, OB. Then Ca, Cb, OA, OB are all perpendicular to OP, because the planes aCb and AOB are perpendicular to OP ; therefore Ca is parallel to OA, and Cb is parallel to OB. Therefore the angle aCb = the angle AOB (Euclid, xi. 10). Hence, nox , Art. 18); therefore, arcab Ca Ca . D - j-^ = -r = -^ = sin POa. arc AB OA Oa II. SPHEEICAL TRIANGLES. 15. Spherical Trigonometry investigates the relations which subsist between the angles of the plane faces which form a solid angle and the angles at which the plane faces are inclined to each other. SPHERICAL TRIANGLES. 7 16. Suppose that the angular point of a solid angle is made the centre of a sphere ; then the planes which form the solid angle will cut the sphere in arcs of great circles. Thus a figure will be formed on the surface of the sphere which is called a spherical triangle if it is bounded by three arcs of great circles ; this will be the case when the solid angle is formed by the meeting of three plane angles. If the solid angle be formed by the meeting of more than three plane angles, the corresponding figure on the surface of the sphere is bounded by more than three arcs of great circles, and is called a spherical polygon. 17. The three arcs of great circles which form a spherical triangle are called the sides of the spherical triangle ; the angles formed by the arcs at the points where they meet are called the angles of the spherical triangle. (See Art. 9.) 18. Thus, let be the centre of a sphere, and suppose a solid angle formed at by the meeting of three plane angles. Let A By J3C, CA be the arcs of great circles in which the planes cut the sphere; then ABC is a spherical triangle, and the arcs AJ3, JBC, CA are its sides. Suppose Ab the tangent at A to the arc AB, and Ac the tangent at A to the arc AC, the tangents being drawn from A towards B and C respectively ; then the angle bAc is one of the angles of the spherical triangle. Similarly angles formed in like manner at B and C are the other angles of the spherical triangle. 8 SPHERICAL TRIANGLES. 19. The principal part of a treatise on Spherical Trigonometry consists of theorems relating to spherical triangles ; it is therefore necessary to obtain an accurate conception of a spherical triangle and its parts. It will be seen that what are called sides of a spherical triangle are really arcs of great circles, and these arcs are pro- portional to the three plane angles which form the solid angle corresponding to the spherical triangle. Thus, in the figure of the preceding Article, the arc AB forms one side of the spherical triangle ABC, and the plane angle AOB is measured by the frac- tion r . 77-7 : and thus the arc AB is proportional to the ano;le radius OA AOB so long as we keep to the same sphere. The angles of a spherical triangle are the inclinations of the. plane faces which form the solid angle ; for since Ab and Ac are both perpendicular to OA, the angle bA c is the angle of inclination of the planes OAB and OAG. 20. The letters A, B, C are generally used to denote the angles of a spherical triangle, and the letters a, b, c are used to denote the sides. As in the case of plane triangles, A, B, and C may be used to denote the numerical values of the angles expressed in terms of any unit, provided we understand distinctly what the unit is. Thus, if the angle C be a right angle, we may say that C = 90, or that C = ^ , according as we adopt for the unit a de- Z gree or the angle subtended at the centre by an arc equal to the radius. So also, as the sides of a spherical triangle are propor- tional to the angles subtended at the centre of the sphere, we may use a, b, c to denote the numerical values of those angles in terms of any unit. We shall usually suppose both the angles and sides of a spherical triangle expressed in circular measure. (Plane Trigonometry, Art. 20.) SPHERICAL TRIANGLES. 9 21. In future, unless the contrary be distinctly stated, any arc drawn on the surface of a sphere will be supposed to be an arc of a great circle. 22. In spherical triangles each side is restricted to be less than a semicircle ; this is of course a convention, and it is adopted because it is found convenient. Thus, in the figure, the arc ADEB is greater than a semicir- cumference, and we might, if we pleased, consider ADEB, AC, and EG as forming a triangle, having its angular points at A, B, and C. But we agree to exclude such triangles from our con- sideration ; and the triangle having its angular points at A, B, and C, will be understood to be that formed by AFB, BC, and CA. 23. From the restriction of the preceding Article it will follow that any angle of a spherical triangle is less than two right angles. For suppose a triangle formed by BO, CA, and BEDA 9 having the angle BCA greater than two right angles. Then suppose D to denote the point at which the arc BC, if produced, will meet AE; then BED is a semicircle by Art. 10, and therefore BE A is greater than a semicircle \ thus the proposedjjriangle is not one of those which we consider. III. SPHERICAL GEOMETRY. 24. The relations between the sides and angles of a Spherical Triangle, which are investigated in treatises on Spherical Trigono- metry, are chiefly such as involve the Trigonometrical Functions of the sides and angles. Before proceeding to these, 'however, we shall collect, under the head of Spherical Geometry, some theorems which involve the sides and angles themselves, and not their trigo- nometrical ratios. 25. Polar triangle. Let ABC be any spherical triangle, and let the points A', B\ C' be those poles of the arcs BC, CA, AB respectively which lie on the same sides of them as the opposite angles A, B, C ; then the triangle A'B'C' is said to be the polar triangle of the triangle ABC. Since there are two poles for each side of a spherical triangle, eight triangles can be formed having for their angular points poles of the sides of the given triangle ; but there is only one triangle in which these poles A', B', C' lie towards the same parts with the corresponding angles A, B, C ; and this is the triangle which is known under the name of the polar triangle. The triangle ABC is called the primitive triangle with respect to the triangle A'B'C'. SPHEEICAL GEOMETRY. 11 26. If one triangle be the polar triangle of another, the latter will be the polar triangle of the former. ~LziABC be any triangle, A'B'C' the polar triangle : then ABC will be the polar triangle of A B'C'. For since B' is a pole of AC, the arc AB' is a quadrant, and since C' is a pole of BA, the arc AC' is a quadrant (Art. 7) ; there- fore A is a pole of B'C' (Art. 11). Also A and A' are on the same side of B'C' ; for A and A' are by hypothesis on the same side of BC, therefore A' A is less than a quadrant ; and since A is a pole of B'C', and A A' is less than a quadrant, J. and J.' are on the same side of B'C'. Similarly it may be shewn that B is a pole of C'A', and that B and B' are on the same side of C'A ; also that C is a pole of J/2?', and that C and (7' are on the same side of A'B'. Thus ABC is the polar triangle of A B'C'. 27. The sides and angles of the polar triangle are respectively the supplements of the angles and sides of the primitive triangle. For let the arc B'C', produced if necessary, meet the arcs AB, AC, produced if necessary, at the points D and E respectively ; then since A is a pole of B'C', the spherical angle A is measured by the arc DE (Art. 12). But B'E arid C'D are each quadrants; therefore DE and B'C' are together equal to a semicircle ; that is, the angle subtended by B'C' at the centre of the sphere is the 12 SPHERICAL GEOMETRY. supplement of the angle A. This we may express for shortness thus ; B'C' is the supplement of A. Similarly it may be shewn that C'A is the supplement of B, and A'E' the supplement of C. And since ABC is the polar triangle of A B'C', it follows that BC, CA, AB are respectively the supplements of A', B' , C r ; that is, A', B', G' are respectively the supplements of BC, CA, AB. From these properties a primitive triangle and its polar tri- angle are sometimes called supplemental triangles. Thus, if A, B, C, a, b, c denote respectively the angles and the sides of a spherical triangle, all expressed in circular measure, and A', B', C', a', b', c' those of the polar triangle, we have A' = TT a, B f = TT b, C' TT c, a' = TT A , b' = TT B, c' TT C. 28. The preceding result is of great importance; for if any general theorem be demonstrated with respect to the sides and the angles of any spherical triangle it holds of course for the polar triangle also. Thus any such theorem will remain true when the angles are changed into the supplements of the corresponding sides and the sides into the supplements of the corresponding angles. "We shall see several examples of this principle in the next Chapter. 29. Any two sides of a spherical triangle are together greater than the third side. (See the figure of Art. 18.) For any two of the three plane angles which form the solid angle at are together greater than the third (Euclid, xi. 20). Therefore any two of the arcs AB, BC, CA, are together greater than the third. From this proposition it is obvious that any side of a spherical triangle is greater than the difference of the other two. 30. The sum of the three sides of a spherical triangle is less than the circumference of a great circle. (See the figure of Art. 18.) SPHERICAL GEOMETRY. 13 For the sum of the three plane angles which form the solid angle at is less than four right angles (Euclid, xi. 21) ; therefore AB EG CA . therefore, AB + EC + CA is less than 2?r x OA ; that is, the sum of the arcs is less than the circumference of a great circle. 31. The propositions contained in the preceding two Articles may be extended. Thus, if there be any polygon which has each of its angles less than two right angles, any one side is less than the sum of all the others. This may be proved by repeated use of Art. 29. Suppose, for example, that the figure has four sides, and let the angular points be denoted by A, B, C, D. Then AB + EG is greater than AC ; therefore, AB + EG + CD is greater than AC + CD, and a fortiori greater than AD. Again, if there be any polygon which has each of its angles less than two right angles, the sum of its sides will be less than the circumference of a great circle. This follows from Euclid, xi. 21, in the manner shewn in Art. 30. 32. The three angles of a spherical triangle are together greater than two right angles and less than six right angles. Let A, B, C be the angles of a spherical triangle j let a', b r , c' be the sides of the polar triangle. Then by Art. 30, a + b f + c' is less than 2?r, that is, TT- A + TT- B + TT (7 is less than 2?r ; therefore, A + B + C is greater than TT. And since each of the angles A, B, C is less than TT, the sum A 4- B + C is less than 3?r. SPHERICAL GEOMETRY. 33. The angles at the base of an isosceles spherical triangle are equal. O Let ABC be a spherical triangle having AC = BC ', let be the centre of the sphere. Draw tangents at the points A and B to the arcs AC and BC respectively; these will meet OC produced at the same point S, and AS will be equal to BS. Draw tangents AT, BT at the points A, B to the arc AB j then AT = TB-, join TS. In the two triangles SAT, SBT the sides SA, AT, TS are equal to SB, BT, TS respectively; therefore the angle SAT is equal to the angle SBT ; and these are the angles at the base of the spherical triangle. The figure supposes AC and BC to be less than quadrants ; if they are greater than quadrants the tangents to AC and BC will meet on CO produced through instead of through C, and the demonstration may be completed as before. If AC and BC are quadrants, the angles at the base are right angles by Arts. 11 and 9. 34. If two angles of a spherical triangle are equal, the opposite sides are equal. Since the primitive triangle has two equal angles, the polar triangle has two equal sides ; therefore in the polar triangle the angles opposite the equal sides are equal by Art. 33. Hence in the primitive triangle the sides opposite the equal angles are equal. SPHERICAL GEOMETRY. 15 35. If one angle of a spherical triangle be greater than ano- ther, the side opposite the greater angle is greater than the side opposite the less angle. Let ABC be a spherical triangle, and let the angle ABC be greater than the angle BAC : then the side AC will be greater than the side BC. At B make the angle ABD equal to the angle BAD ; then BD is equal to AD (Art. 34), and BD + DC is greater than BC (Art. 29) ; therefore AD + DC is greater than BC ; that is, AC is greater than BC. 36. If one side of a spherical triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less side. This follows from the preceding Article by means of the polar triangle. Or thus; suppose the side AC greater than the side BC, then the angle ABC will be greater than the angle BAC. For the angle ABC cannot be less than the angle BAC by Art. 35, and the angle ABC cannot be equal to the angle BAC by Art. 34 ; therefore the angle ABC must be greater than the angle BAC. This Chapter might be extended ; but it is unnecessary to do so because the Trigonometrical formulae of the next Chapter sup- ply an easy method of investigating the theorems of Spherical Geometry. See Arts. 56, 57, and 58. ( 16 ) IY. RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 37. To express the cosine of an angle of a triangle in terms of sines and cosines of the sides. C \ E Let ABC be a spherical triangle, the centre of the sphere. Let the tangent at A to the arc AC meet OC produced at E, and let the tangent at A to the arc AB meet OB produced at D join ED. Thus the angle EAD is the angle A of the spherical triangle, and the angle EOD measures the side a, From the triangles ALE and ODE we have DE* = AD 2 + AE 2 - 2 AD .AficosA, DE 2 = OD 2 + OE 2 -WD. OE cos a ; also the angles OA D and OAE are right angles, so that OD 2 = OA 2 + AD 2 and OE 2 = OA 2 + AE 2 . Hence by subtraction we have = 20 A 2 + 2AD .AEcQsA-WD. OE cos a ; OA OA AE AD therefore cos a = . ^ + ^ . QD cos A ', that is cos a = cos b cos c + sin b sin c cos A t cos a cos b cos c Therefore cos A = - sin b sin c RELATIONS BETWEEN THE FUNCTIONS. 17 38. We have supposed, in the construction of the preceding Article, that the sides which contain the angle A are less than quadrants, for we have assumed that the tangents at A meet OB and OC respectively produced. We must now shew that the formula obtained is true when these sides are not less than quad- rants. This we shall do by special examination of the cases in which one side or each side is greater than a quadrant or equal to a quadrant. (1) Suppose only one of the sides which contain the angle A to be greater than a quadrant, for example, AB. Produce BA and BC to meet at B ; and put AB r c f , CB r = a. Then we have from the triangle AB'C, by what has been already proved, cos a' = cos 1} cos c' + sin b sin c' cos B'AC ; buta' = 7r a, C' = TT-C, B'AC-Tr A-, thus cos a = cos b cos c + sin b sin c cos A. (2) Suppose both the sides which contain the angle A to bo greater than quadrants. Produce AB and AC to meet at A r put A'B c') AC = b'; then from the triangle ABO, as before, cos a = cos b' cos c + sin b' sin c cos A' ; but l'= TT - b, c' = TT - c, A' = A; thus cos a = cos b cos c + sin b sin c cos A. T. S. T. 1 8 KELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS (3) Suppose that one of the sides which contain the angle A is a quadrant, for example, AB on AC, produced if necessary, take AD equal to a quadrant and draw BD. If BD is a quadrant B is a pole of AC (Art. 11) ; in this case a = and A =-^ as well 2t 2 as c = ~ Thus the formula to be verified reduces to the identity 2 = 0. If BD be not a quadrant, the triangle BDC gives cos a = cos (7.Z) cos BD + sin CD sin BD cos C7)j?, and cos CDB = 0, cos (71) = cos (^ ~ 6 J = sin 6, cos .## = cos J. ; thus cos a = sin 6 cos -4 ; and this is what the formula in Art. 37 becomes when c - - . Zi (4) Suppose that both the sides which contain the angle A are quadrants. The formula then becomes cos a = cos A ; and this is obviously true, for A is now the pole of 0, and thus A = a. Thus the formula in Art. 37 is proved to be universally true. 39. The formula in Art. 37 may be applied to express the cosine of any angle of a triangle in terms of sines and cosines of the sides ; thus we have the three formulae, cos a = cos b cos c + sin b sin c cos A, cos 6 = cos c cos a + sin c sin a cos B, cos c - cos a cos b + sin a sin b cos C. OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 19 These may be considered as the fundamental equations of Spheri- cal Trigonometry; we shall proceed to deduce various formulae from them. 40. To express the sine of an angle of a spherical triangle in terms of trigonometrical functions of the sides. cos a cos b cos c We have sin b sin c . /cos a -cos 6 cos c\ 2 therefore sin J A = 11 = 7 ) \ sin 6 sin c J __ (1 cos 8 b) (1 cos 2 c) - (cos a cos b cos c) 2 sin 2 b sin 2 c __ 1 cos 2 a cos 2 & cos 2 c + 2 cos a cos 6 cos c ^ sin 2 6 sure A . - A */(! COS 2 & COS 2 5 COS 2 C + 2 COS COS 5 COSc) therefore sin .4 = 5" : =: . sin b sin c The radical on the right-hand side must be taken with the posi- tive sign, because sin 6, sin c, and sin A are all positive. 41. From the value of sin A in the preceding Article it fol- lows that sin A sin B sin C sin a sin 6 sine ' for each of these is equal to the same expression, namely, fj(l - cos 2 cos 2 6 cos 2 c -f 2 cos a cos b cos c) sin a sin b sin c Thus the sines of the angles of a spJierical triangle are proportional to the sines of the opposite sides. We will give an independent proof of this proposition in the following Article. 42. The sines of tJie angles of a spherical triangle are propor- tional to the sines of the opposite sides. c2 20 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS Let ABC be a spherical triangle, the centre of the sphere. Take any point P in OA, draw PD perpendicular to the plane j and from D draw DE, DF perpendicular to OB, OC respec- tively ; join PE, PF, OD. Since PD is perpendicular to the plane HOC, it makes right angles with every straight line meeting it in that plane ; hence PE* = PD* + DE 2 = PO 2 - OD* + DE 2 - PO 2 - OE 2 ; thus PEO is a right angle. Therefore PE = OP sin POE= OP sin c ; and PD = PE$m PED = PE$mB=OP sin c sin B. Similarly, PD = OP sin b sin C ; therefore OP sin c sin 7? = OP sin 5 sin C ; sin 7? sin b therefore . ~ = . sin C sin c The figure supposes 5, c, j5, and (7 each less than a right angle; it will be found on examination that the proof will hold when the figure is modified to meet any case which can occur. If, for instance, B alone is greater than a right angle, the point D will fall beyond OB instead of between OB and OC ; then PED will be the supplement of B, and thus sin PED is still equal to sin B. 43. To shew that cot a sin 6 = cot A sin (7 + cos 6 cos (7. We have cos a = cos b cos c + sin b sin c cos A, cos c = cos a cos 6 -f sin a sin 6 cos (7, sin O sin c = sin a sin ^4 " twr ^'k OF THE SIDES AND THE ANGLES OP A SPHERICAL TRIANGLE. 21 Substitute the values of cos c and sin c in the first equation ; thus ^ T sin a sin b cos A sin (7 cos a = (cos a cos o + sin a sin 6 cos G) cos 6 + . . ; by transposition cos a sin 2 b = sin a sin b cos 5 cos G + sin a sin b cot J. sin (7 ; divide by sin a sin b ; thus cot a sin 5 = cos b cos (7 -f cot A sin (7. 44. By interchanging the letters five other formula may be obtained like that in the preceding Article ; the whole six formula will be as follows : cot a sin b = cot A sin G + cos b cos G, cot b sin a = cot B sin G + cos a cos C, cot 6 sin c = cot 5 sin A + cos c cos A, cot c sin b = cot (7 sin A + cos 6 cos A, cot c sin a = cot G sin j5 4- cos a cos ^, cot a sin c = cot A sin J5 + cos c cos .Z?. 45. To express the sine, cosine, and tangent, of half an angle of a triangle as functions of the sides. . cos a- cos b cos c We have, by Art. 37, cos A = . , . - : sin b sin c cos a - cos b cos c cos (b - c) - cos a therefore 1 - cos A = 1 : r. = : = -. ; sin 6 sin c sin b sin c . A sin i (a + b - c) sin A (a - b + c) therefore Bin- -^ = ** -I - ' 2 sin 6 sin c Let 2s = a + b + c, so that s is half the sum of the sides of the triangle ; then . J. sin (s 5) sin (s c) thus sin 2 - = - * > 2 sm 6 sin c 22 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS A / ( sin (s - b) sin (s - c)} and . ., cos a cos b cos c cos a cos (b + c) Also, 1 -f cos A - 1 + : = ; - = ; = 7-^ - ; sm 6 sm c sin 6 sin c therefore P J. sin i (& + 5 -f c) sin i (5 -t- c a) sin s sin (5 a) COS 2 = ^ x L z-1 . = . : 2 sin 6 sin c sin 6 sin c ^t /( sin 5 sin (5 -a)*) and cos -^ = A / < : r 4 - x > . 2 \ ( sin b sm c J From the expressions for sin and cos -~ we deduce 2 V 1 sin 5 sin (s - a) / The positive sign must be given to the radicals which occur in A this Article, because -^ is less than a right angle, and therefore its sine, cosine, and tangent are all positive. 46. Since sin A = 2 sin -^ cos -~ , we obtain 2 i sin A = ^. {sin s sin (s - a) sin (s b) sin (s - c)} a . sm b sin c l It may be shewn that the expression for sin.4 in Art. 40 agrees with the present expression by putting the numerator of that expression in factors, as in Plane Trigonometry, Art. 115. We shall find it convenient to use a symbol for the radical in the value of sin A ; we shall denote it by n, so that n 2 = sin s sin (s a) sin (s - b) sin (s c), and 4?i 2 = 1 - cos 2 a - cos 2 6 - eos z c + 2 cos a cos b cos e. OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 23 47. To express the cosine of a side of a triangle in terms of sines and cosines of the angles. In the formula of Art. 37 we may, by Art. 28, change the sides into the supplements of the corresponding angles and the angle into the supplement of the corresponding side ; thus cos (TT-^)=COS (TT--B)cos (TT - C) + sin (TT - B) sin (?r-(7)cos (7r-a), that is, cos A = cos B cos C + sin B sin C cos a. Similarly cos B - cos C cos A + sin C sin A cos b, and cos C = cos A cos .Z? -f sin A sin J5 cos c. 48. The formulae in Art. 44 will of course remain true when the angles and sides are changed into the supplements of the cor- responding sides and angles respectively ; it will be found, how- ever, that no new formulae are thus obtained, but only the same formulae over again. This consideration will furnish some assist- ance in retaining those formulas accurately in the memory. 49. To express the sine, cosine, and tangent, of half a side of a triangle as functions of the angles. cos A + cos B cos C We have, by Art. 47, cos a= - : r-. - - : sin B sin 6 therefore cos A + cos B cos C cos A + cos (B + C) 1 - cos a - : 1 -- : ^. ~ = -- : T, . /T - ; sin B sin G sin B sin 6 , a cos i (A + B + 0) cos therefore sin 2 - = -- ^ - ' - -- - : . -- . 2 sin B sin 6 Let 2S= A+B + C; then B + C- A - 2 (S-A), therefore a_ cos S cos (S A] sin B sin G * cos 8 cos (S -4) ) sinlsinC /' 24 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS - cos A + cos B cos G cos A + cos ( B C) Also 1 + cos a = I + -- . =, . ^ -- = -- ; _ . v '- ; sin B sin G sin Z? sin (7 therefore 2 a _ cos i (A - Jg + (7)cos (A + B-C) _cos (S-E) cos (S-Q >S 2 " sin sin (7 sin sin (7 a /(cos (S- B} cos (- - The positive sign must be given to the radicals which occur in this Article, because ^ is less than a right angle. 50. The expressions in the preceding Article may also be obtained immediately from those given in Art. 45 by means of Art. 28. It may be remarked that the values of sin ? cos ^ , and tan ^ 2t 2t 2i are real. For S is greater than one right angle and less than three light angles by Art. 32 ; therefore cos S is negative. And in the polar triangle any side is less than the sum of the other two ; thus 7T-A is less than Tr-B + ir-G', therefore B + G - A is less than TT; therefore S-A is less than^, and B + G -A is algebraically 2 greater than TT, so that S A is algebraically greater than ^ ; 2 therefore cos (S - A) is positive. Similarly also cos (S - B) and cos (S- G) are positive. Hence the values of sin ^ , cos - , and tan ^ 2i JL 2* are real. 51. Since sin a = 2 sin ^ cos ^ , we obtain 2t Ij 2 sin a = gin ^ g {- cos S cos (tf - ^) cos (S - 5) cos (S - We shall use ^for - OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 25 52. To demonstrate Napier's Analogies. sin A sin B We have . = ; , = m suppose : sin a sin b then, by a theorem of Algebra, sin A + sin B ra= . , (1), sin a + sin b sin A - sin B and also m . 7 ( 2 ). sin a sin b Now cos A + cos B cos C = sin 5 sin G cos a = m sin (7 sin 6 cos a, and cos B -f cos A cos (7 = sin A sin (7 cos b =m sin (7 sin a cos 6, therefore, by addition, (cos J. +cos^)(l + cos (7) = m sin (7 sin (# + 6) (3); therefore by (1) we have sin A + sin B _ sin a + sin b 1 + cos cos J. + cos J5 ~ sin (a + b) sin (7 thatis, tan|(^ + ^) = C08 f / ^"^ooty . ...(4). cos J (a + &) 2 Similarly from (3) and (2) we have sin A sin ^ __ sin a - sin 5 1 + cos C cos A + cos ^ ~~ sin (a + b) sin (7 * that is, tan 1(4 -) = - cot / 6 \ ' sin ^ (a + 6) 2 By writing ?r-4 for a, and so on in (4) and (5) we obtain T . cos i (^ B) J c (7). v x The formulae (4), (5), (6), (7) may be put in the form of pro- portions or analogies, and are called from their discoverer Napier's 26 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS Analogies : the last two may be demonstrated without recurring to the polar triangle by starting with the formulae in Art. 39. 53. In equation (4) of the preceding Article, cos J (a b) and G cot - are necessarily positive quantities ; hence the equation 2t shews that tanJ(J[ +B] and cos J (a + b) are of the same sign; thus ^ ( A + B) and J (a + b) are either both less than a right angle or both greater than a right angle. This is expressed by saying that ^ ( A + B} and J (a + b) are of the same affection. 54:. To demonstrate Delambre's Analogies. We have cos c = cos a cos b + sin a sin b cos C ; therefore 1 + cos c = 1 + cos a cos b + sin a sin b (cos 8 \G sin 2 J C) = (1 + cos (a - b)} cos 2 J (7 + {1 + cos (a + 6)} sin 2 J (7 ; therefore cos 2 1 c = cos 2 J (a - 6) cos 2 J C 4- cos 2 J (a -f b) sin 2 J C. Similarly, sin 2 J c = sin 2 | (a - 6) cos 2 J (7 + sin 2 J (a -f 6) sin 2 J (7. Now add unity to the square of each member of Napier's first two analogies ; hence by the formulae just proved ' 1 / A 7?\ - COs8 ^ C ^ " 2 " ' sn <* + ) sn Extract the square roots ; thus, since J (A + B} and ^ (a 4- &) are of the game affection, we obtain cos J (A +J3) cos |c = cos ^(a + b) sin J (7 ....... (1), cos%(A -j5)sin Jc = sin J (a + 5) sin 1 (7 ....... (2). Multiply the first two of Napier's analogies respectively by these results ; thus sini(^t-f B) cos|c = cos| (a -b) cos \C ....... (3), sin \(A - B) sin ^ c = sin J (a ~ ^) cos 2 ^ ....... W- OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 27 The last four formulae are commonly, but improperly, called Gauss's TJieorems; they were first given by Delambre in the Connaissance des Terns for 1809, page 445. See the Philosophical Magazine for February, 1873. 55. The properties of supplemental triangles were proved geometrically in Art. 27, and by means of these properties the formulse in Art. 47 were obtained; but these formulae may be deduced analytically from those in Art. 39, and thus the whole subject may be made to depend on the formulae of Art. 39. For from Art. 39 we obtain expressions for cos A, cos , cos G\ and from these we find cos A + cos B cos G (cos a cos b cos c) sin* a + (cos b cos a cos c) (cos c cos a cos b) sin 2 a sin b sin c In the numerator of this fraction write 1 cos 2 a for sin 2 a; thus the numerator will be found to reduce to cos a (1 cos* a cos 2 b cos 2 c + 2 cos a cos b cos c), and this is equal to cos a sin B sin C sin 2 a sin b sin c, (Art. 41) ; therefore cos A + cos B cos C = cos a sin .Z? sin C. Similarly the other two corresponding formulae may be proved. Thus the formulae in Art. 47 are established ; and therefore, without assuming the existence and properties of the Polar Tri- angle, we deduce the following theorem : If the sides and angles of a spherical triangle be changed respectively into the supplements of the corresponding angles and sides, the fundamental formulae of Art. 39 hold good, and therefore also all results deducible from them. 56. The formulae in the present Chapter may be applied to establish analytically various propositions respecting spherical tri- angles which either have been proved geometrically in the pre- ceding Chapter, or may be so proved. Thus, for example, the second of Napier's analogies is 1 . . sin A (a b) C tan \ (A -B) =. . f ; J cot - 6 ; 7 sin A (a + b} 2 ' 28 KELATIONS BETWEEN THE TRIGONOMETKICAL FUNCTIONS this shews that J ( J. - J3) is positive, negative, or zero, according as J (a b) is positive, negative, or zero ; thus we obtain all the re- sults included in Arts. 33... 36. 57. If two triangles have two sides of the one equal to two sides of the other, each to each, and likewise the included angles equal, then their other angles will be equal, each to each, and like- wise their loses will be equal. We may shew that the bases are equal by applying the first formula in Art. 39 to each triangle, supposing b, c, and A the same in the two triangles; then the remaining two formulae of Art. 39 will shew that B and G are the same in the two triangles. It should be observed that the two triangles in this case are not necessarily such that one may be made to coincide with the other by superposition. The sides of one may be equal to those of the other, each to each, but in a reverse order, as in the following figures. Two triangles which are equal in this manner are said to be symmetrically equal ; when they are equal so as to admit of super- position they are said to be absolutely equal. 58. If two spherical triangles have two sides of the one equal to two sides of the other , each to each, but the angle which is contained by the two sides of the one greater than the angle which is contained by the two sides which are equal to them of the other, the base of that OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 29 which has the greater angle will be greater than the base of the other ; and conversely. Let b and c denote the sides which are equal in the two tri- angles ; let a be the base and A the opposite angle of one triangle, and a and A' similar quantities for the other. Then cos a = cos b cos c + sin b sin c cos A, cos a' = cos b cos c + sin b sin c cos A' ; therefore cos a cos a = sin b sin c (cos A cos A' ) ; that is, sin \ (a -f a') sin J (a - a') = sin b sin c sin | (A + .4') sin ^ (A ^t') ; this shews that J (a a') and J (J. A') are of the same sign. 59. If on a sphere any point be taken within a circle which is not its pole, of all the arcs which can be drawn from that point to the circumference, the greatest is that in which the pole is, and the other part of that produced is the least ; and of any others, that which is nearer to the greatest is always greater than one more remote; and from the same point to the circumference there can be drawn only two arcs which are equal to each other, and these make equal angles with the shortest arc on opposite sides of it. This follows readily from the preceding three Articles. 60. We will give another proof of the fundamental formulee in Art. 39, which is very simple, requiring only a knowledge of the elements of Co-ordinate Geometry. Suppose ABC any spherical triangle, the centre of the sphere, take as the origin of co-ordinates, and let the axis of z pass through C. Let x^ y^ z t be the co-ordinates of A, and x a , y 2 , z y those of B let r be the radius of the sphere. Then the square on the straight line AB is equal to (a, -*.)'+ fa- ?.)*+(*, -O'. and also to r 5 + r* - 2r 5 cos A OB ; 30 EXAMPLES. and x* + y* + z* - r 2 , x* + y? -f z 2 z = r 2 , thus Now make the usual substitutions in passing from rectangular to polar co-ordinates, namely, z l = r cos 00 oij = r sin } cos c^, y l = r sin O l sin c^, 2 2 = r cos 2 , # a = r sin 2 cos < 2 , 2/ 2 = r sin 2 sin <^> 2 ; thus we obtain cos 2 cos } + sin ^ 2 sin } cos (^> l <^> 2 ) = cos AOJ3, that is, in the ordinary notation of Spherical Trigonometry, cos a cos b + sin a sin b cos (7 = cos c. This method has the advantage of giving a perfectly general proof, as all the equations used are universally true. EXAMPLES. 1. If A =a, shew that B and b are equal or supplemental, as also C and c. 2. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle. 3. When does the polar triangle coincide with the primitive triangle 1 4. If D be the middle point of AB, shew that cos AC + cos = 2 cos \ AB cos CD. 5. If two angles of a spherical triangle be respectively equal to the sides opposite to them, shew that the remaining side is the supplement of the remaining angle ; or else that the triangle has two quadrants and two right angles, and then the remaining side is equal to the remaining angle. EXAMPLES. 31 6. In an equilateral triangle, shew that 2 cos sin = 1. I. In an equilateral triangle, shew that tan 2 -= 1 2 cos A; A hence deduce the limits between which the sides and the angles of an equilateral triangle are restricted. 8. In an equilateral triangle, shew that sec A = 1 + sec a. 9. If the three sides of a spherical triangle be halved and b c a new triangle formed, the angle between the new sides -^ and ^ >j l/ is given by cos = cos A + J tan ^ tan ^ sin 2 0. 2t 2i 10. AB, CD are quadrants on the surface of a sphere inter- secting at E, the extremities being joined by great circles : shew that cos AEG = cos AC cos BD - cos BG cos AD. II. If b + c = TT, shew that sin 2B + sin 20 = 0. 12. If DE be an arc of a great circle bisecting the sides AB 9 AC of a spherical triangle at D and ^, P a pole of D^, and P.Z?, .PZ), PJ&, P(7 be joined by arcs of great circles, shew that the angle BPG = twice the angle DPE. 13. In a spherical triangle shew that sin b sin c + cos b cos c cos ^ = sin B sin (7 cos B cos (7 cos a. 14. If D be any point in the side BC of a triangle, shew that cos AD sin 5(7 = cos AB sin _D(7 + cos AC sin -Z?Z>. 15. In a spherical triangle shew that if 0, <, i/r be the lengths of arcs of great circles drawn from A, B, C perpendicular to the opposite sides, sin a sin = sin b sin - sin c sin \f/ = V (1 cos 2 a - cos 2 b - cos 2 c + 2 cos a cos# cos c). 32 SOLUTION OF RIGHT-ANGLED TRIANGLES. 16. In a spherical triangle, if 0, <, if/ be the arcs bisecting the angles A, B, C respectively and terminated by the opposite sides, shew that ABC cot cos ^ + cot < cos 4- cot if/ cos = cot a + cot b + cot c. Jj Z 2t 17. Two ports are in the same parallel of latitude, their com- mon latitude being I and their difference of longitude 2X : shew that the saving of distance in sailing from one to the other on the great circle, instead of sailing due East or West, is 2r {X cos I sin" 1 (sin X cos Z)}, X being expressed in circular measure, and r being the radius of the Earth. 18. If a ship be proceeding uniformly along a great circle and the observed latitudes be Z 1? l a l z , at equal intervals of time, in each of which the distance traversed is 8, shew that _ x sin i (/ + 7 ) cos \ (I - 1) - S-iJ - - 2LU - a/ r denoting the Earth's radius : and shew that the change of longi- tude may also be found in terms of the three latitudes. V. SOLUTION OF EIGHT-ANGLED TRIANGLES. 61. In every spherical triangle there are six elements, namely, the three sides and the three angles, besides the radius of the sphere, which is supposed constant. The solution of spherical tri- angles is the process by which, when the values of a sufficient number of the six elements are given, we calculate the values of the remaining elements. It will appear, as we proceed, that when the values of three of the elements are given, those of the remain- ing three can generally be found. We begin with the right-angled triangle : here two elements, in addition to the right angle, will be supposed known. SOLUTION OF RIGHT-ANGLED TRIANGLES. 33 62. The formulae requisite for the solution of right-angled triangles may be obtained from the preceding Chapter by sup- posing one of the angles a right angle, as for example. They may also be obtained veiy easily in an independent manner, as we will now shew. Let A EG be a spherical triangle having a right angle at G ; let be the centre of the sphere. From any point P in OA draw PM perpendicular to 0(7, and from M draw MN perpendicular to OB, and join PN. Then PM is perpendicular to MN, because the plane AOC is perpendicular to the plane BOG ; hence PN 2 = PM 2 + MN 2 = OP 2 - OM 2 + OM 2 - ON 2 = OP 2 - ON 2 -, therefore PNO is a right angle. And ON ON OM PM_PM PN Similarly sin a = sin A sin c I MN MN PN ..,,., ) - , that is, tana = cos B tan c f ........ (3), Similarly tan b = cos A tan c ' PM PM MN , - A -^' tliatlS?tan ^ tan ^ Sina l ........ W- Similarly tan a = tan ^ sin 6 ) T. S. T, 34 SOLUTION OF RIGHT-ANGLED TRIANGLES. Multiply together the two formulae (4) ; .thus, D tan a tan b 1 1 tan A tan B = . , - = = by (1) ; sm a sin b cos a cos b cos c ' therefore cos c = cot A cotB (5). Multiply crosswise the second formula in (2) and the first in (3) ; thus sin a cos B tan c = tan a sin A sin c ; therefore cos B = = sin A cos b by (1). cos a Thus cos B = sin A cos 6| Similarly cos A = sin ^ cos a ) " These six formulae comprise ten equations; and thus we can solve every case of right-angled triangles. For every one of these ten equations is a distinct combination involving three out of the five quantities a, 6, c, A y B\ and out of five quantities only ten combinations of three can be formed. Thus any two of the five quantities being given and a third required, some one of the pre- ceding ten equations will serve to determine that third quantity. 63. As we have stated, the above six formulae may be ob- tained from those given in the preceding Chapter by supposing C a right angle. Thus (1) follows from Art. 39, (2) from Art. 41, (3) from the fourth and fifth equations of Art. 44, (4) from the first and second equations of Art. (44), (5) from the third equation of Art. 47, (6) from the first and second equations of Art. 47. Since the six formulae may be obtained from those given in the preceding Chapter which have been proved to be universally true, we do not stop to shew that the demonstration of Art. 62 may be applied to every case which can occur ; the student may for exercise investigate the modifications which will be necessary when we suppose one or more of the quantities a, b, c, A, B equal to a right angle or greater than a right angle. SOLUTION OF EIGHT-ANGLED TRIANGLES, 35 64. Certain properties of right-angled triangles are deducible from the formulae of Art. 62. From (1) it follows that cos c has the same sign as the product cos a cos b ; hence either all the cosines are positive, or else only one is positive. Therefore in a right-angled triangle either all the three sides are less than quadrants, or else one side is less than a quadrant and the other two sides are greater than quadrants. From (4) it follows that tana has the same sign as tan A Therefore A and a are either both greater than - , or both less than -^ ; this is expressed by saying that A and a are of the same 2i affection. Similarly B and b are of the same affection. 65. The formulae of Art. 62 are comprised in the following enunciations, which the student will find it useful to remember ; the results are distinguished by the same numbers as have been already applied to them in Art. 62 ; the side opposite the right angle is called the hypotenuse : Cos hyp = product of cosines of sides (1), Cos hyp = product of cotangents of angles ...., (5), Sine side = sine of opposite angle x sine hyp .(2), Tan side = tan hyp x cos included angle (3), Tan side = tan opposite angle x sine of other side (4), Cos angle= cos opposite side x sine of other angle (6). 66. Napier's Rules. The formulae of Art. 62 are comprised in two rules, which are called, from their inventor, Napier's Rules of Circular Parts. Napier was also the inventor of Logarithms, and the Rules of Circular Parts were first published by him in a work entitled Mirifici Logarithmorum Canonis Descriptio Edinburgh, 1614. These rules we will now explain. D2 36 SOLUTION OF EIGHT-ANGLED TRIANGLES. The right angle is left "out of consideration ; the two sides which include the right angle, the complement of the hypotenuse, and the complements of the other angles are called the circular parts of the triangle. Thus there are Jive circular parts, namely, a, b, TT A, ^ c, #; and these are supposed to be ranged 2t 21 2S round a circle in the order in which they naturally occur with respect to the triangle. Any one of the five parts may be selected and called the middle part, then the two parts next to it are called adjacent parts, and the remaining two parts are called opposite parts. For example, if ^ B is selected as the middle part, then the adjacent A parts are a and c, and the opposite parts are b and ^ A. Then Napier's Rules are the following : sine of the middle part = product of tangents of adjacent parts, sine of the middle part = product of cosines of opposite parts. 67. Napier's Rules may be demonstrated by shewing that they agree with the results already established. The following table shews the required agreement : in the first column are given the middle parts, in the second column the results of Napier's Rules, and in the third column the same results expressed as in Art. 62, with the number for reference used in that Article. SOLUTION OF EIGHT- ANGLED TRIANGLES. 37 ?- o sin (? - c) = tan fc - A\ tan (| - JB\ cos c = cot A cot ..(5), sin(^-cj = cosa cos 6 cos c = cos a cos &..(!), in ( H~^) - tana tan (?-c) cos J5 = tan a cote ..(3), in-^J = cos b cos (J^-A\ cos .5 = cos 5 sin J..(6), sin a = tan b tan ( - B ) sin a = tan 6 cot ^..(4), sin a = cos \^-A\ cos (H- C ) sin a = sin A sine. .(2), -- sn H b = tan ( A j tan a sin 6 = cot A tan a.. (4), sin b = cos f--^ cos f- f- cj sin b = sin B sine.. (2), - A sin ( A } = tan b tan ( - c J cos A = tan 6 cot c .. (3), sinf ^ -4 j = cos a cos (-^ B\ cos ^1 = cos a sin .#..(6), The last four cases need not have been given, since it is obvious that they are only repetitions of what had previously been given ; the seventh and eighth are repetitions of the fifth and sixth, and the ninth and tenth are repetitions of the third and fourth. 68. It has been sometimes stated that the method of the preceding Article is the only one by which Napier's Rules can be demonstrated ; this statement, however, is inaccurate, since besides this method Napier himself indicated another method of proof in his Mirifici Logarithmorum Canonis Descriptio, pp. 32, 35. This we will now briefly explain, 38 SOLUTION OF EIGHT-ANGLED TRIANGLES. Let ABC be a spherical triangle right-angled at (7; with B as pole describe a great circle DEFG, and with A as pole describe a great circle HFKL, and produce the sides of the original triangle ABC to meet these great circles. Then since B is a pole of DEFG the angles at D and G are right angles, and since A is a pole of HFKL the angles at H and L are right angles. Hence the five triangles BAG, AED, EFH, FKG, KBL are all right-angled; and moreover it will be found on examination that, although the ele- ments of these triangles are different, yet their circular parts are the same. We will consider, for example, the triangle AED ; the angle EAD is equal to the angle BAC ; the side AD is the com- plement of AB ; as the angles at C and G are right angles E is a pole of GC (Art. 13), therefore EA is the complement of AC \ as B is a pole of DE the angle BED is a right angle, therefore the angle AED is the complement of the angle BEC, that is, the angle AED is the complement of the side BC (Art. 12); and simi- larly the side DE is equal to the angle DBE, and is therefore the complement of the angle ABC. Hence, if we denote the elements of the triangle ABC as usual by a, &, c, A, B, we have in the triangle AED the hypotenuse equal to ^-6, the angles equal to A and ^-a, and the sides respectively opposite these angles equal 2i to*-fi and ^-c. The circular parts of AED are therefore the 2 2 SOLUTION OF RIGHT-ANGLED TRIANGLES. 39 same as those of ABC. Similarly the remaining three of the five right-angled triangles may be shewn to have the same circular parts as the triangle ABC has. Now take two of the theorems in Art. 60, for example (1) and (3); then the truth of the ten cases comprised in Napier's Rules will be found to follow from applying the two theorems in succes- sion to the five triangles formed in the preceding figure. Thus this method of considering Napier's Rules regards each Rule, not as the statement of dissimilar properties of one triangle, but as the statement of similar properties of five allied triangles. 69. In Napier's work a figure is given of which that in the preceding Article is a copy, except that different letters are used ; Napier briefly intimates that the truth of the Rules can be easily seen by means of this figure, as well as by the method of induction from consideration of all the cases which can occur. The late T. S. Davies, in his edition of Dr Hutton's Course of Mathematics, drew attention to Napier's own views and expanded the demon- stration by a systematic examination of the figure of the preceding Article. It is however easy to evade the necessity of examining the whole figure ; all that is wanted is to observe the connexion between the triangle AED and the triangle BAG. For let a l9 a 2 , a 3 j a^ a 5 represent the elements of the triangle BAC taken in order, beginning with the hypotenuse and omitting the right angle; then the elements of the triangle AED taken in order, beginning with the hypotenuse and omitting the right angle, are o ~~ a & ~c> ~" a 4> <> ~~ a & o ~~ a i> an d a a 9 -^? therefore, to characterise L 2i ' 2i 2t the former we introduce a new set of quantities p l9 p s , p a , p 4 , p 59 such that a l + p^ = a 2 + p a = a 5 +p. = , and that p 3 = a a and p 4 = 4 , then the original triangle being characterised by p l9 p s , p^ p^ p S9 the second triangle will be similarly characterised by p a9 p^ p., p l9 p 2 . As the second triangle can give rise to a third in like manner, and so on, we see that every right-angled triangle is one 40 SOLUTION OF EIGHT-ANGLED TRIANGLES. of a system of five such triangles which, are all characterised by the quantities p l9 p^ p 3 , p 4 , p s , always taken in order, each quantity in its turn standing first. The late R. L. Ellis pointed out this connexion between the five triangles, and thus gave the true significance of Napier's Rules. The memoir containing Mr Ellis's investigations, which was unpublished when the first edition of the present work ap- peared, will be found in pages 32 8... 335 of The Mathematical and other writings of Robert Leslie Ellis... Cambridge, 1863. Napier's own method of considering his Rules was neglected by writers on the subject until the late T. S. Davies drew atten- tion to it. Hence, as we have already remarked in Art. 68, an erroneous statement was made respecting the Rules. For in- stance, Woodhouse says, in his Trigonometry : "There is no sepa- rate and independent proof of these rules;..." Airy says, in the treatise on Trigonometry in the Encyclopaedia, Metropolitana : " These rules are proved to be true only by showing that they com- prehend all the equations which we have just found." 70. Opinions have diftered with respect to the utility of Napier's Rules in practice. Thus Woodhouse says, "In the whole compass of mathematical science there cannot be found, perhaps, rules which more completely attain that which is the proper object of rules, namely, facility and brevity of computation." (Trigonometry, chap, x.) On the other hand may be set the fol- lowing sentence from Airy's Trigonometry (Encyclopcedia Metro- politana)'. "In the opinion of Delambre (and no one was better qualified by experience to give an opinion) these theorems are best recollected by the practical calculator in their unconnected form." See Delambre's Astronomic, vol. I. p. 205. Professor De Morgan strongly objects to Napier's Rules, and says (Spherical Trigono- metry, Art. 17): "There are certain mnemonical formulae called Napier's Rules of Circular Parts, which are generally explained. We do not give them, because we are convinced that they only create confusion instead of assisting the memory." SOLUTION OF EIGHT-ANGLED TRIANGLES. 41 71. We shall now proceed to apply the formulae of Art. 62 to the solution of right-angled triangles. We shall assume that the given quantities are subject to the limitations which are stated in Arts. 22 and 23, that is, a given side must be less than the semicircumference of a great circle, and a given angle less than two right angles. There will be six cases to consider. 72. Having given the hypotenuse c and an angle A. Here we have from (3), (5) and (2) of Art. 62, tan b = tan c cos A, cot B = cose tan A, sin a sine sin A. Thus b and B are determined immediately without ambiguity ; and as a must be of the same affection as A (Art. 64), a also is determined without ambiguity. It is obvious from the formulae of solution, that in this case the triangle is always possible. If c and A are both right angles, a is a right angle, and b and B are indeterminate. 73. Having given a side b and the adjacent angle A, Here we have from (3), (4) and (6) of Art. 62, tan c = 7 , tan a = tan A sin b. cos B = cos b sin A. cos A Here c, a, B are determined without ambiguity, and the tri- angle is always possible. 74. Having given the two sides a and b. Here we have from (1) and (4) of Art. 62, cos c = cos a cos 6, cot A = cot a sin b, cot B = cot b sin a. Here c, A, B are determined without ambiguity, and the tri- angle is always possible. 75. Having given the hypotenuse c and a side a. Here we have from (1), (3) and (2) of Art. 62, . cos c tan a . sin a cos b = , cos B = . sin A = - , cosa tan c sin c 42 SOLUTION OF RIGHT-ANGLED TRIANGLES. Here b, B, A are determined without ambiguity, since A must be of the same affection as a. It will be seen from these formulae that there are limitations of the data in order" to insure a possible triangle ; in fact, c must lie between a and tr a in order that the values found for cos 6, cos -5, and sin A may be numerically not greater than unity. If c and a "are right angles, A is a right angle, and b and B are indeterminate. 76. Having given the two angles A and B. Here we have from (5) and (6) of Art. 62, , n cos A t cosfi cos c = cot A cot B. cos a = ^ , cos b = -. . sin B sin A Here c, a, b are determined without ambiguity. There are limitations of the data in order to insure a possible triangle. First suppose A less than ^ , then B must He between - -A and ^ + A; 2t Jj A next suppose A greater than - , then B must lie between 2 -^ (TT A) and -f (TT ^4), that is, between A ^ and A. 2t 2i 2i 2t 77. Having given a side a and the opposite angle A. Here we have from (2), (4) and (6) of Art. 62, sin a cos A sin c = -r r , sin b = tan a cot A, sin B = . sin J. cos a Here there is an ambiguity, as the parts are determined from their sines. If sin a be less than sin .4, there are two values admissible for c; corresponding to each of these there will be in general only one admissible value of b, since we must have cos c = cos a cos 6, and only one admissible value of B, since we must have cos c = cot A cot B. Thus if one triangle exists with the given parts, there will be in general two, and only two, triangles with the given parts. We say in general in the preceding sen- tences, because if a = A there will be only one triangle, unless a SOLUTION OF RIGHT-ANGLED TRIANGLES. 43 and A are each right angles, and then b and B become inde- terminate. It is easy to see from a figure that the ambiguity must occur in general. For, suppose BAG to be a triangle which satisfies the given conditions ; produce AB and AC to meet again at A' then the triangle A'BC also satisfies the given conditions, for it has a right angle at C, BC the given side, and A' = A the given angle. If a = A, then the formulae of solution shew that c, 5, and B are right angles ; in this case A is the pole of BC, and the triangle ABC is symmetrically equal to the triangle ABC (Art. 57). If a and A are both right angles, B is the pole of AC j B and b are then equal, but may have any value whatever. There are limitations of the data in order to insure a possible triangle. A and a must have the same affection by Art. 64; hence the formulae of solution shew that a must be less than A if both are acute, and greater than A if both are obtuse. EXAMPLES. If A BC be a triangle in which the angle C is a right . angle, prove the following relations contained in Examples 1 to 5. ~. O c . O a 9 b Q cb . b 1. Sin 2 -= = sin 2 - cos 2 ^ + cos 2 ^ sin 2 - . ..' 2i 2i 2i A 2. T^n |(c + a) tan l(c - a) = tan 2 - . ; "J_ 3. Sin (c - b) = tan 2 sin (c + b). 44 EXAMPLES. 4. Sin a tan \A -sin 6 tan^J5 = sin (a-b). 5. Sin (c a) sin b cos a tan J B, Sin (c a) = tan b cos c tan J Z?. 6. If ABC be a spherical triangle, right-angled at 0, and cos A= cos 2 a, shew that if A be not a right angle & + C = |TT or 3 7T - TT, according as b and c are both less or both greater than - , 2 A 7. If a, /? be the arcs drawn from the right angle respectively perpendicular to and bisecting the hypotenuse c, shew that M sin 2 - (1 + sin 2 a) = sin 2 ft. t 8. In a triangle, if be a right angle and D the middle point of AB, shew that 4 cos 2 sin 2 CD = sin 2 a + sin* b. 2i 9. In a right-angled triangle, if 8 be the length of the arc drawn from C perpendicular to the hypotenuse AB, shew that cot 8= 10. OAA l is a spherical triangle right-angled at A I and acute- angled at A y the arc A^ z of a great circle is drawn perpendicular to OA, then A 2 A 3 is drawn perpendicular to OA l9 and so on : shew that A n A n+l vanishes when n becomes infinite; and find the value j cos A^A of these circles cut off by great circles through the point, and through two points on the axes, each - from their 2 point of intersection : shew that if the three points (0, <), (0', <'), (0", <") lie on the same great circle tan < (tan ff - tan 0") + tan <' (tan 0" - tan 0) + tan <" (tan - tan 0') = 0. 17. If a point on a sphere be referred to two great circles at right angles to each other as axes, by means of the portions of these axes cut off by great circles drawn through the point and two points on the axes each 90 from their intersection, shew that the equation to a great circle is tan cot a + tan < cot ft = 1. 18. In a spherical triangle, if A = - , B = - 9 and C = ^ , shew o 6 2 that a + b + c = ^ . J VI. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 78; The solution of oblique-angled triangles may be made in some cases to depend immediately on the solution of right-angled triangles } we will indicate these cases before considering the sub- ject generally. (1) Suppose a triangle to have one of its given sides equal to a quadrant. In this case the polar triangle has its corresponding "angle a right angle ; the polar triangle can therefore be solved by the rules of the preceding Chapter, and thus the elements of the primitive triangle become known. (2) Suppose among the given elements of a triangle there are two equal sides or two equal angles. By drawing an arc from the vertex to the middle point of the base, the triangle is divided into two equal right-angled triangles ; by the solution of one of these right-angled triangles the required elements can be found. (3) Suppose among the given elements of a triangle there are two sides, one of which is the supplement of the other, or two angles, one of which is the supplement of the other. Suppose, for example, that b + C = TT, or else that B + C = TT ; produce BA and EC to meet at B' (see the first figure to Art. 38); then the triangle B'AC has two equal sides given, or else two equal angles given; and by the preceding case the solution of it can be made to depend on the solution of a right-angled triangle. 79. We now proceed to the solution of oblique-angled tri- angles in general. There will be six cases to consider. 80. Having given the three sides. cos a cos i Here we have cos A = : JT- sin o sn for cos B and cos (7. Or if we wish to use formulae suited to loga- cos a cos b cos c , . ., Here we have cos A = . 7 -. , and similar formulae sin b sin c SOLUTION OF OBLIQUE-ANGLED TEIANGLES. 47 rithms, we may take the formula for the sine, cosine, or tangent of half an angle given in Art. 45. In selecting a formula, attention should be paid to the remarks in Plane Trigonometry, Chap. XII. towards the end. 81. Having given the three angles. cos A + cos B cos C , . ., Here we have cosa = -- . . ~~ , and similar formulae sin B sin 6 for cos b and cos c. Or if we wish to use formulae suited to loga- rithms, we may take the formula for the sine, cosine, or tangent of half a side given in Art. 49. There is no ambiguity in the two preceding cases; the triangles however may be impossible with the given elements. 82. Having given two sides and the included angle (a, C, b). By Napier's analogies - . . . cos i (a - b) tan i (A + B) = - f-} - ^ -cot i C, ' cos \ (a + 6) 2 sin i (a b) n - . sin (a + these determine ^ (A + B] and \ (A ^), and thence -4 and B. Then c may be found from the formula sin c = - : - : in sin .4 this case, since c is found from its sine, it may be uncertain which of two values is to be given to it ; the point may be sometimes settled by observing that the greater side of a triangle is opposite to the greater angle. Or we may determine c from equation (1) of Art. 54, which is free from ambiguity. Or we may determine c, without previously determining A and B, from the formula cos c = cos a cos b -r sin a sin b cos C \ this is free from ambiguity. This formula may be adapted to logarithms thus : cos c = cos b (cos a + sin a tan b cos C) ; 48 SOLUTION OF OBLIQUE-ANGLED TRIANGLES, assume tan = tan b cos C ; then , , . m cos b cos (a - 0) cos c = cos o (cos a + sin a tan U) = ^ ' : COS0 this is adapted to logarithms. Or we may treat this case conveniently by resolving the tri- angle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to GB or GB produced ; then, by Art. 62, tan CD = tan b cos (7, and this determines CD, and then DB is known. Again, by Art. 62, cos c = cos AD cos DB = cos DB cos b this finds c. It is obvious that CD is what was denoted by in the former part of the Article, By Art. 62, tan AD = tan C sin CD, and tan AD= tan ABD sin DB ; thus tan ^.#Z) sin DB = tan (7 sin 0, where DB = a 6 or a, according as D is on GB or (7Z? pro- duced, and ABD is either B or the supplement of B; this for- mula enables us to find B independently of A. Thus, in the present case, there is 110 real ambiguity, and the triangle is always possible. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 49 83. Having given two angles and the included side (A, c, B). By Napier's analogies, i / T\ cosi(^-J9) x 1 tan i (a + b) = - f-f-j - ^ tan I c, 7 cos^(^l + .#) J 3 7 x sin & (A B) , .. these determine (a + b) and |(a b), and thence a and 6. sin J. sin c Then (7 may be found from the formula sm G = - ; - : in sin a this case, since C is found from its sine, it may be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater angle of a triangle is opposite to the greater side. Or we may determine C from equation (3) of Art. 54, which is free from ambiguity. Or we may determine C without previously determining a and b from the formula cos C = cos A cos B + sin A sin B cos c. This formula may be adapted to logarithms, thus : cos C = cosB ( cos A + sin A tan B cos c) ; assume cot < = tan B cos c ; then n T> / , t \ cos -B sin (A ) cos G= cos B ( cos A + cot < sin A) = sin this is adapted to logarithms. Or. we may treat this case conveniently by resolving the tri- angle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to CB (see the right- hand figure of Art. 82) ; then, by Art. 62, cos c = cot B cot DAB, and this determines DAB, and then CAD is known. Again, by Art. 62, cos AD sin CAD = cos C, and cos AD sin BAD = cos B ; cos C cos B ,. . _ _ _. therefore -: = -5 . ^ : this finds C. srnCAD sin. BAD' It is obvious that DAB is what was denoted by in the former part of the Article. T. S. T. E 50 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. By Art. 62, tan AD = tan AC cos CAD, and tan AD = tan AB cos BAD ; thus tan b cos (LIZ) = tan c cos <, where CAD = A - < ; this formula enables us to find b indepen- dently of a. Similarly we may proceed when the perpendicular AD falls on CB produced ; (see the left-hand figure of Art. 82). Thus, in the present case, there is no real ambiguity; more- over the triangle is always possible. 84. Having given two sides and the angle opposite one of them (a, b, A). The angle B may be found from the formula . _ sin 6 . . sin B - - sin A : sin a and then C and c may be found from Napier's analogies, cos ^ (a + b) In this case, since B is found from its sine, there will sometimes be two solutions ; and sometimes there will be no solution at all, namely, when the value found for sin B is greater than unity. We will presently return to this point. (See Art. 86.) We may also determine C and c independently of B by for- mulae adapted to logarithms. For, by Art. 44, cot a sin b = cos b cos C + sin C cot A = cos b (cos C + -- j- sin C) ; cot A assume tan 9 = - j- ; tnus ^ x cos b cos (C 6) cot a sin b = cos b (cos C + tan < sin C) = - -jr - - ; COS

is to be found, and then C. The ambi- guity still exists ; for if the last equation leads to C <> = a, it will be satisfied also by < - C = a ; so that we have two admissible values for C, if < + a is less than TT, and < a is positive. And cos a = cos b cos c + sin b sin c cos A cos b (cos c + sin c tan b cos A) ; assume tan = tan 5 cos A ; thus cos 6 cos (c 0) cos a = cos 6 (cos c + sin c tan 0) = - COS0 therefore from this equation c is to be found, and then c ; and there may be an ambiguity as before. Or we may treat this case conveniently by resolving the tri- angle into the sum or difference of two right-angled triangles. Let CA = b, and let CAE = the given angle A ; from C draw CD perpendicular to AE, and let CB and CB 1 = a; thus the figure shews that there may be two triangles which have the given ele- ments. Then, by Art. 62, cos b = cot A cot ACD j this finds ACD. Again, by Art. 62, tan CD = tan AC cos ACD 9 and tan CD = tan CB cos BCD, or tan CB' cos BCD, therefore tan AC cos ACD = tan CB cos ^(7Z>, or tan CB cos .g'(7Z) ; this finds BCD or J5 r (7Z>. It is obvious that ACD is what was denoted by < in the former part of the Article. E2 52 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. Also, by Art. 62, tan AD = tan AC cos A ; this finds AD. Then cos AC = cos CD cos AD, cos CB = cos CD cos J5Z>, or cosCB' = co$CDcosB'D; cos .4 (7 cos CB cos (7JS' therefore A tl = -- 57-, or -- ^-yr cos AD cos .#/> cos -D D this finds J5Z> or B'D. It is obvious that AD is what was denoted by $ in the former part of the Article. 85. Having given two angles and the side opposite one of them (A, B, a). This case is analogous to that immediately preceding, and gives rise to the same ambiguities. The side b may be found from the formula . 7 sin B sin a sm 6 = - : -j : sin A and then C and c may be found from Napier's analogies, ton coB^(a-5) co cos \ (a + b) cos , tan J c = -=( tan J (a + b). * - We may also determine C and c independently of b by formulae adapted to logarithms. For cos A = cos -5 cos C + sin -5 sin C cos a = cos B ( cos 6' + tan J5 sin (7 cos a), assume cot < = tan B cos a ; thus cos A = cos B (- cos (7 + sin C cot <)= - ^i ""^J; therefore sin (C - <) = -- _ - ; cos .5 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 53 from, this equation C < is to be found and then C. Since C < is found from its sine there may be an ambiguity. Again, by Art. 44, L A - T> 7i T> / cot a sin c\ cot A sin/? = cot a sin c - cos c cos J5 = cos B { - cos c + ) , \ cos B J X /) C0t a XT. assume cot B ^ : then cos B L A r r> / . , COS J? sin (c - 0) cot J. siri B = cos jo ( cos c 4- sin c cot 6) = ; : sin 6 therefore sin (, and ft and ft' are both greater than A, and both inadmissible. Hence there is no solution. II. Let b be equal to ^ . 2 (1) Let a be less than b; then ft and ft' are both greater than Ay and both admissible. Hence there are two solutions. (2) Let a be equal to b ; then there is no solution, as pre- viously shewn. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 55 (3) Let a be greater than b ; then sin a is less than sin 5, and ft and ft' are both greater than A 9 and inadmissible. Hence there is no solution. III. Let b be greater than ^ , (1) Let a be less than 5; we may have then a + b less than TT or equal to TT or greater than TT. If a + b is less than TT, then sin a is less than sin 6, and ft and ft' are both greater than A and both admissible. Hence there are two solutions. If a + b is equal to -TT, then ft is equal to A and inadmissible, and ft f is greater than -4 and admissible. Hence there is one solution. If a + b is greater than TT, then sin a is greater than sin b ; ft is less than .4 and inadmissible, and ft' is greater than A and admissible. Hence there is one solution. (2) Let a be equal to b ; then there is no solution, as pre- viously shewn. (3) Let a be greater than 5; then sin a is less than sin 6, and ft and ft? are both greater than A and both inadmissible. Hence there is no solution. "We have then the following results when A is less than a right angle. Ia b and a + b < TT ........................... one solution, I a > b and a + b = TT or > TT .................. no solution. TT ( a < b ..................... ......................... two solutions, _ .' 2 (a = 6ora>6 ....... , ......................... no solution. I a < b and a + 6<7r ........................... two solutions, a 7r .................. one solution, a = bor>b . .............. no solution. 56 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. It must be remembered, however, that in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A. In the same manner the cases in which A is equal to a right angle or greater than a right angle may be discussed, and the following results obtained. When A is equal to a right angle, (a < b or a = b 110 solution, a>b and a + b b and a + & = 7ror>7r no solution. ,_T (ab no solution, 2 \a = b infinite number of solutions. !a < b and a + b>7r one solution, a < b and a + b = TT or < TT no solution, a~ b or a>6 no solution. When A is greater tfian a right angle, ^ (a 6 and a + 6 = TT or < TT. one solution, (a > 6 and a -f 5 > TT two solutions. , _TT Ca b v . two solutions. a < b and a+b>7r one solution, a < b and a + 6 = TT or < TT no solution, 6 one solution, a> b two solutions. As before in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A. It will be seen from the above investigations that if a lies between b and TT b, there will be one solution ; if a does not lie between b and ?r b either there are two solutions or there is no solution ; this enunciation is not meant to include the cases in which a = b or = TT b. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 57 87. The results of the preceding Article may be illustrated by figure. Let ADA'E be a great circle; suppose PA and PA' the projections on the plane of this circle of arcs which are each equal to b and inclined at an angle A to ADA'; let PD and PE be the projections of the- least and greatest distances of P from the great circle (see Art. 59). Thus the figure supposes A and b each less than - . 2 If a be less than the arc which is represented by PD there is no triangle ; if a be between PD and PA in magnitude, there are two triangles, since B will fall on ADA', and we have two triangles BPA and BPA ; if a be between PA and PH there will be only one triangle, as B will fall on AH or AH', and the triangle will be either APB with B between A' and H, or else A'PB with B be- tween A and H' ; but these two triangles are symmetrically equal (Art. 57) ; if a be greater than PH there will be no triangle. The figure will easily serve for all the cases ; thus if A is greater than ^ , we can suppose PAE and PAE to be equal to A ; if A b is greater than ^ , we can take PH and PH to represent 6. 58 SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 88. The ambiguities which, occur in the last case in the solu- tion of oblique-angled triangles (Art. 85) may be discussed in the same manner as those in Art. 86 ; or, by means of the polar triangle, the last case may be deduced from that of Art. 86. EXAMPLES. 1. The sides of a triangle are 105, 90, and 75 respectively : find the sines of all the angles. 2. Shew that tan J -4 tan J 2? = '. Solve a triangle when a side, an adjacent angle, and the sum of the other two sides are given. 3. Solve a triangle having given a side, an adjacent angle, and the sum of the other two angles. 4. A triangle has the sum of two sides equal to a semicir- cumference : find the arc joining the vertex with the middle of the base. 5. If a, b, c are known, c being a quadrant, determine the angles : shew also that if 8 be the perpendicular on c from the opposite angle, cos 8 8 = cos 2 a 4- cos 2 b. 6. If one side of a spherical triangle be divided into four equal parts, and O l9 2 , 3 , 4 be the angles subtended at the oppo- site angle by the parts taken in order, shew that sin (6 l -f 2 ) sin 2 sin 4 = sin (0 3 + 4 ) sin O l sin 3 . 7. In a spherical triangle if A = B = 2(7, shew that / c\ . s c c . 8 8 sin ( a + Q \ sin 2 - cos -^ = sin a. CIRCUMSCRIBED AND INSCRIBED CIRCLES. 59 8. In a spherical triangle if A = B = 2(7, shew that C* f C*\ ^ 8 sin 2 ( cos s + sin -^ ) -=!. 2\ 2/cosa 9. If the equal sides of an isosceles triangle ABC be bisected by an arc DE, and BC be the base, shew that sin ^=1 sin sec- 1 ? 2 ~~ 2 2 fc ' '2 " 10. If c 1? c a be the two values of the third side when A, a, b are given and the triangle is ambiguous, shew that c c tan - 1 tan - = tan \ (b - a) tan J (b + a). VII. CIRCUMSCRIBED AND INSCRIBED CIRCLES. 89. To find the angular radius of tlw small circle inscribed in a given triangle. Let ABC be the triangle ; bisect the angles A and B by arcs meeting at P ; from P draw PJ), PE, PF perpendicular to the sides. Then it may be shewn that PD, PJE, PF are all equal ; also that AE = AF, BF=BD, CD = CE. Hence BC + AF= half the sum of the sides = 5; therefore AF = s a. Let PF= r. Now tan PF= tan PAFsin AF (Art. 62) ; thus tan r - tan sin (s - a) (1). CO CIRCUMSCRIBED AND INSCRIBED CIRCLES. The value of tan r may be expressed in various forms ; thus from Art. 45, we obtain t an 4. - //sin(g-5)sin(*--c)) ; 2 V 1 sin s sin (s - a) y substitute this value in (1), thus tanr= /(^( 8 -a)sin(.-6)sin( g -c)) = _ VI sins J sins v Again sin (* - a) = sin {| (b + c) - J a} = sin J (6 + c) cos J a - cos J (b + c) sin J a sin , ( cos H 5 - c^) ~ cos sin a sin ^ ^ sin | C ,, /. /1X , . therefore from (1) tan r = ^TA sma COS A. hence, by Art. 51, 2 cos ^ -4 cos ^ j5 cos ^ (7 "~2cos^cosJ.ecos<7" It may be shewn by common trigonometrical formulae that 4 cos I A cos JjtfcosJC^ cos S + cos(S-A) + cos^-JBJ+cos^-C'); hence we have from (4) cot r = [costf + cos (S-A) + cos (S-S) + cos (S- C)\ (5). CIRCUMSCRIBED AND INSCRIBED CIRCLES. 61 90. To find the angular radius of the small circle described so as to touch one side of a given triangle, and the other sides produced. Let ABC be the triangle; and suppose we require the radius of the small circle which touches BC, and AB and AC produced. Produce AB and AC to meet at A '; then we require the radius of the small circle inscribed in A'BC, and the sides of A f BC are a, TT b, TT c respectively. Hence if r l be the required radius, and * denote as usual ^ (a -4- b + c), we have from Art. 89, tan r tan sin 5 .............................. (1). 2 From this result we may derive other equivalent forms as in the preceding Article ; or we may make use of those forms im- mediately, observing that the angles of the triangle A! EG are A, ir-B, TT C respectively. Hence 5 being (a + b -f c) and 8 being J (A + B + C) we shall obtain sin s sin (s - b) sin (s - -- s^> cos ^ B cos i C . = - (3), x ; ' tanr = N/{~ cos ^ cos (^"^) cos (^-- g ) cos (#- ^ cos J ^ sin ^ B sin 6' ^ cos cot r t = J- cos /?- cos (5 - ^) + cos (S-B) + cos (tf- C)], . .(5). G2 CIRCUMSCRIBED AND INSCRIBED CIRCLES. These results may also be- found independently by bisecting two of the angles of the triangle A'BC, so as to determine the pole of the small circle, and proceeding as in Art. 89. 91. A circle which touches one side of a triangle and the other sides produced is called an escribed circle; thus there are three escribed circles belonging to a given triangle. We may denote the radii of the escribed circles which touch CA and AB respectively by r f and r a , and values of tanr a and tanr 3 may be found from what has been already given with respect to tan^ by appropriate changes in the letters which denote the sides and angles. In the preceding Article a triangle A 'EC was formed by pro- ducing AB and AC to meet again at A '; similarly another triangle may be formed by producing BC and BA to meet again, and another by producing CA and CB to meet again. The original triangle ABC and the three formed from it have been called associated triangles, ABC being the fundamental triangle. Thus the inscribed and escribed circles of a given triangle are the same as the circles inscribed in the system of associated triangles of which the given triangle is the fundamental triangle. 92. To find the angular radius of the small circle described about a given triangle. Let ABC be the given triangle ; bisect the sides CB, CA at D and E respectively, and draw from D and E arcs at right angles to CB and CA respectively, and let P be the intersection of these Xl CIRCUMSCRIBED AND INSCRIBED CIRCLES. 03 arcs. Then P will be the pole of the small circle described about ABC. For draw PA, PB, PC ; then from the right-angled triangles PCD and PBD it follows that PB = PC' y and from the right-angled triangles PCE and P^^ it follows that PA = P6' ; hence PA=PB = PC. Also the angle P^^ = the angle P#J:, the angle PBC = the angle PCB, and the angle PC A = the angle ; therefore PCB + A = J (4 + ^ + <7), and Now tan CD = tan C7P cos P<7Z>, (Art. 62), thus tan ^ a = tan -S cos (S -4), therefore tan R = **?$".. ... ...(I). cos(S A) The value of tan R may be expressed in various forms j thus if we substitute for tan - from Art. 49, we obtain A _ _ - _ ___ Vlcos(AS r -^)cos(AS v -^)cos(AS f -(7)J ' N ' Again cos (S - A) = cos {| (B + C) - \ A} = cos %(B + C) cos J ^1 + sin ^(B + C) sin -J ^1 sin A- ^t cos A , ,, x - ,. x , /. J (ft + c) + cos I (6 - c)}, (Art. 54 5 ) sin .4 , , . - T -cospcosi C ; therefore from ( 1 ) _ 6 cos c Substitute in the last expression the value of sin .4 from Art. 46 ; thus f r> 2 sin |- a sin J & sin ^ c ^/{sin s sin (s a) sin (s b) sin (s c)} 2 sin J a sin J b sin .(4). 64 CIRCUMSCRIBED AND INSCRIBED CIRCLES. It may be shewn, by common trigonometrical formula?, that 4 sin J a sin b sin -| c = sin (s - a) + sin (5 - ft) + sin (s - c) sin 5 ; hence WQ have from (4) ........... (5). 93. To find the angular radii of the small circles described round the triangles associated with a given fundamental triangle. Let H l denote the radius of the circle described round the triangle formed by producing AB and AG to meet again at A'; similarly let JR 2 and R 3 denote the radii of the circles described round the other two triangles which are similarly formed. Then we may deduce expressions for tanJ?^ tan7? 2 , and tan R a from those found in Art. 92 for tan R. The sides of the triangle A'BC are a, TT b, TT c, and its angles are A, TT B, irC ] hence if s = (a + b + c) and S = ^ (A + B + C} wo shall obtain from Art 92 _ tan ^ a tan E -*-a .... fl) /( cos^-^i) ^ cos(^-^) 1 ~ V \ cos > S r cos(AS'-^)cos(AS r -(7] ))- N -^ 7? sn a tan/? = - j : fr-i T ......................................... (3), sin -d sin J 6 sin ^ c 2 sin J- a cos 5 cos i c =-77-= - : 7-^ , x 7 . , - r , .................. (4), ^{sin s sin (s - a) sin (s - 6) sm (s-c)} = -jsins -sin(s a) + sin(5-i) + sin ( c)> ......... (5). Similarly we may find expressions for tan E a and tan 7i' 3 . 94. Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles. "We will give one that will be of use hereafter. CIRCUMSCRIBED AND INSCRIBED CIRCLES. 65 To prove that (cot r + tan JR)* = i (sin a + sin6 + sin c) 2 - 1. We have 4n* = 1 cos 2 a cos 2 b cos* c -f 2 cos a cos 6 cos c ; therefore (sin a + sin 6 4- sin c) 2 M = 2 ( 1 4- sin a sin 5 -f- sin b sin c -f sin c sin a cos a cos 5 cos c j . Also cot r 4- tan .72 = ^- ! sin s + sin (s a) + sin (s 6) + sin (s c) > ; and by squaring both members of this equation the required result will be obtained. For it may be shewn by reduction that sin 2 s + sin 2 (s a) + sin 2 (s b) -f sin 2 (s c) = 2 2 cos a cos 6 cos c, and sin s sin (s a) + sin s sin (5 &) 4- sin s sin (5 c) -f sin (5 a) sin (s b) + sin (s b) sin (5 - c) + sin (s c) sin (s - a) = sin a sin 6 + sin 6 sin c + sin c sin a. Similarly we may prove that (cot r l tan jff) 2 = -r z (sin b + sin c sin a) 2 - 1. 95. In the figure to Art. 89, suppose DP produced through P to a point A' such that DA' is a quadrant, then J/ is a pole of J5(7, and PA! = -rj similarly, suppose EP produced through P to a point B r such that -fi'.Z?' is a quadrant, and ^P produced through P to a point C" such that FC r is a quadrant. Then A'ffC' is the polar triangle of ABC, and PA' = PB' = PC f = ^-r. A Thus P is the pole of the small circle described round the polar triangle, and the angular radius of the small circle described round the polar triangle is the complement of the angular radius of the T. S. T. F 66 CIRCUMSCRIBED AND INSCRIBED CIRCLES. small circle inscribed in the primitive triangle. And in like man- ner the point which is the pole of the small circle inscribed in the polar triangle is also the pole of the small circle described round the primitive triangle, and the angular radii of the two circles are complementary. EXAMPLES, In the following examples the notation of the Chapter is retained. Shew that in any triangle the following relations hold con- tained in Examples 1 to 7 : 1 . Tan r l tan r a tan r a = tan r sin 2 8. 2. Tan R + cot r = tan R l + cot r l = tan R a + cot r a = tan 7? 3 + cot r 3 = (cot r + cot r l + cot r a + cot rj. 3. Tan 2 R + tan 2 E l + tan 2 R a + tan 2 R 3 = cot 2 r + cot 2 r l + cot 2 r a + cot 2 r 3 . Tan r, + tan r. + tan r. - tan r - ._ x 4. l =i(l + cosa-f coso + CGSC). cot r l + cot r 2 + cot r 3 cot r 5. CosecV= cot (s-a) cot (s-b) + cot (s-b) cot (s-c) 4- cot (s-c)(s-a). 6. Cosec 2 r x = cot (s b) cot (s - c) cot s cot (5 b) cot s cot (s - c). 7 . Tan R 1 tan # a tan J? 3 = tan # sec 2 S. 8. Shew that in an equilateral triangle tan R = 2 tan r. 9. If ABC be an equilateral spherical triangle, P the pole of the circle circumscribing it, Q any point on the sphere, shew that cos QA + cos QB + cos QG = 3 cos 7\4 cos P(). 10. If three small circles be inscribed in a spherical triangle having each of its angles 120, so that each touches the other two as well as two sides of the triangle, shew that the radius of each of the small circles = 30, and that the centres of the three small circles coincide with the angular points of the polar triangle. VIII. AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 96. To find the area, of a Lune. A Lune is that portion of the surface of a sphere which is com- prised between two great semicircles. Let ACBDA, AD BE A be two lunes having equal angles at A ; then one of these lunes may be supposed placed on the other so as to coincide exactly with it; thus lunes having equal angles are equal. Then by a process similar to that used in the first propo- sition of the Sixth Book of Euclid it may be shewn that lunes are proportional to their angles. Hence since the whole surface of a sphere may be considered as a lune with an angle equal to four right angles, we have for a lune with an angle of which the circular measure is A, area of lune A surface of sphere 2ir * Suppose r the radius of the sphere, then the surface is 4?rr 2 (Integral Calculus, Chap, vn.) ; thus area of lune = ^ 4?rr 2 = 2Ar 3 . JTT 68 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 97. To find the area of a SpJierical Triangle. Let AEG be a spherical triangle ; produce the arcs which form its sides until they meet again two and two, which will happen when each has become equal to the semi-circumference. The triangle ABC now forms a part of three lunes, namely, ABDCA, BCEAB, and CAFBC. Now the triangles CDE and AFB are subtended by vertically opposite solid angles at 0, and we will assume that their areas are equal ; therefore the lune CAFBC is equal to the sum of the two triangles ABC and CDE. Hence if A 9 B, C denote the circular measures of the angles of the triangle, we have triangle ABC + BGDC = lune ABDCA = 2Ar*, triangle ABO + AHEC = lune BCEAB = 2Br*, triangle ABC + triangle CDE = lune CAFBC = 2Cr* ; hence, by addition, twice triangle ABC + surface of hemisphere = 2 (A + B 4- C) r 3 ; therefore triangle ABC = (A + B + C - 7r)r 2 . The expression A+B + C - TT is called the spherical excess of the triangle; and since Q AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 69 the result obtained may be thus enunciated : the area of a spherical triangle is the same fraction of half the surface of the sphere as the spherical excess is of four right angles. 98. We have assumed, as is usually done, that the areas of the triangles CDE and AFE in the preceding Article are equal. The triangles are, however, not absolutely equal, but symmetri- cally equal (Art. 57), so that one cannot be made to coincide with the other by superposition. It is, however, easy to decom- pose two such triangles into pieces which admit of superposition, and thus to prove that their areas are equal. For describe a small circle round each, then the angular radii of these circles will be equal by Art. 92. If the pole of th circumscribing circle falls inside each triangle, then each triangle is the sum of three isosceles triangles, and if the pole falls outside each triangle, then each triangle is the excess of two isosceles triangles over a third ; and in each case the isosceles triangles of one set are respectively absolutely equal to the corresponding isosceles triangles of the other set. 99. To find the area of a spherical polygon. Let n be the number of sides of the polygon, 2 the sum of all its angles. Take any point within the polygon and join it with all the angular points ; thus the figure is divided into n triangles. Hence, by Art. 97, area of polygon = (sum of the angles of the triangles - mr) r a , and the sum of the angles of the triangles is equal to S together with the four right angles which are formed round the common vertex; therefore area of polygon = < 2 (n 2) TT > r*. This expression is true even when the polygon has some of its angles greater than two right angles, provided it can be decom- posed into triangles, of which each of the angles is less than two right angles. 70 AHEA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 100. We shall now give some expressions for certain trigono- metrical functions of the spherical excess of a triangle. We denote the spherical excess by H, so that E = A+B + C-ir. 101. CagnoWs Theorem. To shew that . r ^/{sin s sin (5 a) sin (s b) sin (s c)} * Sin -R JLJ = - - * - - *= - = - -- . 2 cos J a cos J 6 cos J c i ( a _ 5) _ cos i ( a + ft)}, (Art. 54), - COS "2" C sin (7 sin J a sin J 6 cos Jc sin i a sin A- 6 2 ,, . . , \ / T\ / \> - s - 2- . _ -- : _ . J{sm8 sm (5- a) sin (5-6) sin (5- c) cos ^ c sin a sm 6 ^/{sin s sin (s a) sin (5 - b) sin (5 c)} 2 cos J a cos \ b cos J c 102. Lhuilier's Theorem. To shew that tan J j = ^/{tan 1 5 tan |- (5 - a) tan | (5 - &) tan J (s - c)}. sin i (ir C) /Tt7 m . \, Q4 \ - n 2V^F, Art. 84), - - - , cos %(A + B) + cos J (?r - C) 7 _cosj-(a-5)-coslc cos_JC (Art. 54). cos J (a + b) + cos | c ' sin J C ' Hence, by Art. 45, we obtain sin^(c + a-6)sin^(c-f b-a) /( sin s sin (g - c) ) =r cosi(a+6 + c)cosi(a-l-6-c)\/ |sin(s-a)sin(5-6)/ = ,/{tan ^5 tan J (* - a) tan J (*-&) tan J (s -c}}. AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 71 103. "We may obtain many other formula involving trigo- nometrical functions of the spherical excess. Thus, for example, cos = cos = cos J (A + B) sin J G + sin %(A + B) cos J C cos ^(a + b) sin 2 |(7 4- cos|(a - &) cos 2 |-(7 i sec Jc, (Art. 54), = -j = \ cos J a cos J 5 (cos 2 J (7 + sin 2 J C) + sin J a sin J 6 (cos 2 J (7 - sin 2 J (7) I sec J c = {cos Ja cos J5 -f sin J a sin J6 cos C} sec Jc, ........ (1). Again, it was shewn in Art. 101, that sin J E = sin C sin J a sin J b sec J c ; . sin |a sin 16 sin (7 therefore tan*^= - = =^ - r~ -. n - 77 ..... (2). cos Jet cos Jo + sin Ja sin Jo cos 6 Again, we have from above = -I cos J a cos |- 5 + sin J a sin J 6 cos C > sec J c cos _ (1 + cos a) (1 + cos b) + sin a sin b cos (7 4 cos J a cos J 6 cos J c _ 1 + cos a + cos o 4- cos c _ cos 2 \a + cos 2 J6 + cos* \c I 4 cos J a cos J b cos J c ~~ .2 cos J a cos J 6 cos J c In (3) put 1-2 sin 2 |^ for cos J E ; thus ' 2 1 W ^ + ^ COS i a COS 2 b COS I" C "~ COs2 \ a ~' COs2 i ~ COs2 2" C 4 cos J a cos J 6 cos J c By ordinary development we can shew that the numerator of the above fraction is equal to 4 sin J s sin J (s - a) sin |(s - b) sin J (s - c) ; 72 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. therefore _ sinjs sin J (s - a) s in|(g -o) sin J (g-c) m bill -rMf T~i 1 . l:rl cos J a cos J o cos J c Similarly . m j E, _ COS J S COS J(g - g) COS J (g - 5) COS J(g - C) ^Uo ^r JLV .j _ ,*.('.// cos Jacos Jo cos Jc Hence by division we obtain Lhuilier's Theorem. Again, rfnft7-UF) - \ 1 g, 7 = sin C cot i^- cos (7 sin J.# ~ cos i a cos i 6 4- sin 4 a sin i 5 cos C ~ , /Ck . = sin - r-4 - . i 2 , . ^ 2 --- cos (7, by (2), sm ^ a sin J 6 sin (7 = cot J a cot Jo; therefore, by Art. 101, N/( sin g sin ( g ~ a ) sin ( g - ty sin ( g " c )} sin (C - - ^^ 2 sin Ja sin Jo cos J c Again, cos (C ^E) = cos (7 cos J^ + sin C sin ~- r^ - 1 +sin C/sinJasiniosecic 4 cos J a cos J b cos J c (1 + cos a) (1 4- cos b) cos C + sin a sin o 4 cos J a cos J 6 cos J c = ! cos J a cos J 6 cos C + sin Ja sin J 6 > sec J c sin a sin 5 cos (7 + 4 sin 2 Ja sin 2 J 6 4 sin J a sin J 6 cos J c _ cos c cos a cos 6 + (1 cos a) (1 cos 6) 4 sin J a sin J 6 cos J c _ 1 + cos c cos a cos 6 _ cos 2 Jc cos 2 Ja - cos 2 J o + 1 ^ 4 sin Jasin Jo cos Jc " 2 sin Jasin Jo cos J c AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 73 From this result we can deduce two other results, in the same manner as (4) and (5) were deduced from (3) ; or we may observe that the right-hand member of (6) can be obtained from the right-hand member of (3) by writing TT a and TT b for a and b respectively, and thus we may deduce the results more easily. We shall have then sin \ a sin J b cos J c cos'Q-C in^ 4 ' sin ^ a sin ^ 6 cos ^ c EXAMPLES. 1. Find the angles and sides of an equilateral triangle whose area is one-fourth of that of the sphere on which it is described. 2. Find the surface of an equilateral and equiangular sphe- rical polygon of n sides, and determine the value of each of the angles when the surface equals half the surface of the sphere. 3. If a = 6 = - , and c = ^ , shew that E = cos" 1 ^ . o 2t y 4. If the angle C of a spherical triangle be a right angle, shew that sin | E= sin J a sin | b sec J c, cos J E- cos | a cos J 6 sec | c. 5. If the angle (7 be a right angle, shew that sin 2 c _, sin 2 a sin 2 6 cos E = + r . cos c cos a cos 6 6. If a = b and O = , shew that tan .# = J!H^ . 2 2 cos a 7. The sum of the angles in a right-angled triangle is less than four right angles. 8. Draw through a given point in the side of a spherical triangle an arc of a great circle cutting off a given part of the triangle. 74 EXAMPLES. 9. In a spherical triangle if cos C = -tan ^ tan-, then 10. If the angles of a spherical triangle be together equal to four right angles cos 8 \ a 4- cos 2 1- b + cos 2 \ c 1. 11. If r,, r 2 , r 3 be the radii of three small circles of a sphere of radius r which touch one another at P, Q, J?, and A, B, C be the angles of the spherical triangle formed by joining their centres, area PQR = (A cos r l + B cos r a + C cos r 3 - TT) r*. 12. Shew that jsin I #sin (4 - J^) sin (5- 1^) sin ((7- sins = 2 sin J J. sin J B sin (7 13. Given two sides of a spherical triangle, determine when the area is a maximum. 14. Find the area of a regular polygon of a given number of sides formed by arcs of great circles on the surface of a sphere ; and hence deduce that, if a be the angular radius of a small circle, its area is to that of the whole surface of the sphere as versin a is to 2. 15. A 9 B, C are the angular points of a spherical triangle; A', B', C' are the middle points of the respectively opposite sides. If E be the spherical excess of the triangle, shew that coaA'ff cos-S'tf' cos CM' COS J j& = -- i - = -- -, = - 5-7- cos J c cos a cos J 6 16. If one of the arcs of great circles which join the middle points of the sides of a spherical triangle be a quadrant, shew that the other two are also quadrants. IX. ON CERTAIN APPEOXIMATE FORMULAE. 104. We shall now investigate certain approximate formulae which are often useful in calculating spherical triangles when the radius of the sphere is large compared with the lengths of the sides of the triangles. 105. Given two sides and the included angle of a spherical triangle, tafind the angle between the chords of these sides. Let AE, AC be the two sides of the triangle AEG', let be the centre of the sphere. Describe a sphere round A as a centre, and suppose it to meet AO, AB, AC at D, E, F respectively. Then the angle ELF is the inclination of the planes OAJB, OAC, and is therefore equal to A. From the spherical triangle DEF cos EF= cos DE cos DF+ sin DE sin DF cos A ; and therefore cos EF ** sin \ b sin |- c + cos J b cos \ c cos A. If the sides of ' the triangle are small compared with the radius of the sphere, EF will not differ much from A- suppose EF=A 0, then approximately cos EF= cos A + 6 sin A ; 70 ON CERTAIN APPROXIMATE FORMULA. and sin J b sin J c = sin 2 J (6 + c) - sin 8 \ (b - c), cos 6 cos J c = cos 2 J (6 + c) sin 2 (6 - c) ; therefore cos A -f 6 sin J. = sin 2 J (6 + c) - sin 2 (6 - c) 4- 1 1 - sin 2 \ (b -f c) - sin* J (6 - c) j cos 4 ; therefore therefore = tan 4 sin 2 (6 + c)-cot J. sin 2 J(6-c). This gives the circular measure of ; the number of seconds in the angle is found by dividing the circular measure by the circular measure of one second, or approximately by the sine of one second (Plane Trigonometry, Art. 123). If the lengths of the arcs corre- sponding to a and b respectively be a and /?, and r the radius of the sphere, we have - and - as the circular measures of a and b r r respectively ; and the lengths of the sides of the chorda! triangle are 2rsin^- and 2rsin^- respectively. Thus when the sides of the spherical triangle and the radius of the sphere are known, we can calculate the angles and sides of the chordal triangle. 106. Legendre's Theorem. If the sides of a spherical triangle be small compared with the radius of the sphere, then each angle of the spherical triangle exceeds by one third of the sp/ierical ex- cess the corresponding angle of the plane triangle, the sides of which are of the same length as the arcs of the spherical triangle. Let A, B, be the angles of the spherical triangle ; a, b, c the sides ; r the radius of the sphere ; a, (3, y the lengths of the arcs which form the sides, so that -, -, - are the circular r r r measures of a, b, c respectively. Then ON CERTAIN APPROXIMATE FORMULAE. 77 cos a cos b cos c cos A = - sin b sin c - a 2 now cos a = 1 7^-5 3 6r Similar expressions hold for cos b and sin 6, and for cos c and sin c respectively. Hence, if we neglect powers of the cir- cular measure above the fourth, we have Kow let -4', B', C' be the angles of the plane triangle whose sides are a, /?, y respectively ; then thus Suppose A = A' + 0; then cos -4 = cos A' - sin A' approximately ; therefore 78 ON CERTAIN APPROXIMATE FORMULA. where S denotes the area of the plane triangle whose sides are a, /?, y. Similarly B=B' + ~, and (7 = C' + p; hence approximately A+B + C = A'+ B' + C' + - = * + -,; r 2 r } nr therefore t is approximately equal to the spherical excess of the spherical triangle, and thus the theorem is established. It will be seen that in the above approximation the area of the spherical triangle is considered equal to the area of the plane triangle which can be formed with sides of the same length. 107. Legendre's Theorem may be used for the approximate solution of spherical triangles in the following manner. (1) Suppose the three sides of a spherical triangle known; then the values of a, /?, y are known, and by the formulae of Plane Trigonometry we can calculate S and A', B', C' ; then A 9 B 9 C arecknown from the formulae ' ' ' (2) Suppose two sides and the included angle of a spherical triangle known, for example A, b, c. Then S = \ (3y sin A ' = |- fly sin A approximately. cr Then A' is known from the formula A'= A ^ . Thus in the plane triangle two sides and the included angle are known : therefore its remaining parts can be calculated, and then those of the spherical triangle become known. ON CERTAIN APPROXIMATE FORMULAE. 79 (3) Suppose two sides and the angle opposite to one of them in a spherical triangle known, for example A, a, b. Then /? /? sin B f - - sin A f - - sin A approximately ; a a and C'-ir A' B'=TT A B r approximately; then S= \ a/3 sin C'. Hence A' is known and the plane triangle can be solved, since two sides and the angle opposite to one of them are known. (4) Suppose two angles and the included side of a spheri- cal triangle known, for example, A, B, c. ' sin B r y 2 sin A sin B Hence in the plane triangle two angles and the included side are known. (5) Suppose two angles and the side opposite to one of them in a spherical triangle known, for example A, B, a. Then (7 = TT A' B' = TT A B, approximately, and ~ a 2 sin B* sin C' which can be calculated, since B' and C' are approximately known. 108. The importance of Legendre's Theorem in the applica- tion of Spherical Trigonometry to the measurement of the Earth's surface has given rise to various developments of it which enable us to test the degree of exactness of the approximation. We shall finish the present Chapter with some of these developments, which will serve as exercises for the student. We have seen that ap- o proximately the spherical excess is equal to 2 , and we shall begin with investigating a closer approximate formula for the spherical excess. 80 ON CERTAIN APPROXIMATE FORMULA. 109. To find an approximate value of the spherical excess. Lot E denote the spherical excess ; then sin i a sin A 6 sin C = - ? -- r -; cos c therefore approximately therefore and sin C = sin (7 + 1 = sin C' From (1) and (2) Hence to this order of approximation the area of the spheri- cal triangle exceeds that of the plane triangle by the fraction a **j*** y * of the latter. 2 110. To find an approximate value of -- . Sin A sin a , . , , sin A x hence approximately ^ = ' ON CERTAIN APPROXIMATE FORMULA. 81 120r 4 6r 2 36r 4 120r* 36r 4 / 0r~ r ~6^ + TO r - a'-fl 4 F(fP-a?)) * 4 36r 4 j 111, T 7 ^ express cot 5 cot ^1 approximately. Cot J5 cot ^1 = -: =r (cos B . . cos A) : sm j5 x sin J. hence, approximately, by Art. 110, 1 /? fia 2 -p 2 cot 2r- cot ^1 =-^ _ (cos 5 - - cos A - -- -- ' cos A ). sm-5 x a a 6r 2 Now we have shewn in Art. 106, that approximately ff + y - a' a' + F + / - 2a*/3* - 2j8y - 2y V -~ O ~2 _ O2 therefore cos 2? - - cos A = - approximately, a ay - - -- and cot B - cot ^1 = - -^ --- ^- D ^ . ay sin B ay sin /) 1 2r ay sin ^ 112. The approximations in Arts. 109 and 110 are true so far as terms involving r 4 ; that in Art. Ill is true so far as terms involving r 2 , and it will be seen that we are thus able to carry the approximations in the following Article so far as terms involving ? >4 . T. S. T. G 82 ON CERTAIN APPROXIMATE FORMULAE. 113. To find an approximate value of the error in the length of a side of a spherical triangle when calculated by Legendrds Theorem. Suppose the side /? known and the side a required ; let 3/x de- note the spherical excess which is adopted. Then the approximate valued T^ - ^-is taken for the side of which a is the real sin (B p.) value. Let x =a- ~ - r* we nave then to find x ap- sm (B - p) proxirnately. Now approximately M 2 . , . N sin A - p. cos A - ^- sin A sm (A - p) r 2 sin(j8-/x)~ ft 8 . ^' sm B p. cos B - sin E 2 ^ jl -f p. (cot B - cot ^l) + ^ cot B (cot B - cot A)\ sin A asm A - = - n - sm B sm B Also the following formulae are true so far as terms involv- ing r 2 : sin A _ a / ft 2 - a 2 " ~" sin A _ a / ft 2 - a 2 \ sin"^ ~ ^ \ ~6^"/ ' Hence, approximately, Therefore x = a ON CERTAIN APPROXIMATE FORMULAE. 83 /3smA ft (a 2 -/3 s ) -. -.--- D sin B y sin B If we calculate p. from the formula p. = ^ !, 2 we obtain g(/?*-a 2 )(3a*~7/3 2 ) 3607-* If we calculate p. from an equation corresponding to (1) of Art. 109, we have 1-f or i ~ /*_* therefore 720^ MISCELLANEOUS EXAMPLES. 1. If the sides of a spherical triangle AB, AC be produced to B', (7', so that BB r , CC' are the semi-supplements of AB, AC respectively, shew that the arc B'C' will subtend an angle at the centre of the sphere equal to the angle between the chords of AB and AC. 2. Deduce Legendre's Theorem from the formula A sin A (a + b c) sin i (c + a b) tan = =-2 - -* . 2 sin | (6 + c - a) sin J (a + b + c) 3. Four points A, B, C, D on the surface of a sphere are joined by arcs of great circles, and U, F are the middle points of the arcs AC, BD : shew that cos AB + cos BC + cos CD + cos DA = 4 cos AE cos HFcos FE. 4. If a quadrilateral A BCD be inscribed in a small circle on a sphere so that two opposite angles A and C may be at opposite extremities of a diameter, the sum of the cosines of the sides is constant. 84 MISCELLANEOUS EXAMPLES. 5. In a spherical triangle if A = B = 20, shew that a / a\ cos a cos y - cos f c + ^ J . 6. ABC is a spherical triangle each of whose sides is a quad- rant ; P is any point within the triangle : shew that cos PA cos PB cos PG + cot BPC cot CPA cot APE = 0, and tan ABP tan (7P tan CXP = 1. 7. If be the middle point of an equilateral triangle ABC, and P any point on the surface of the sphere, then J (tan PO tan OA ) 2 (cos P4 + cos P -f cos PC) 2 = cos 2 PA+cos 2 PB+cos 2 PC-cosPAcoaPB-cosPBcosPC-cosPCcosPA. 8. If ABC be a triangle having each side a quadrant, the pole of the inscribed circle, P any point on the sphere, then (cos PA + cos PB + cos PC) 2 = 3 cos 2 PO. 9. From each of three points on the surface of a sphere arcs are drawn on the surface to three other points situated on a great circle of the sphere, and their cosines are a,b,c; a', b f , c' ; a", b", c". Shew that ab"c + a'bc" + a"b'c = ab'c" + a'b"c + a"bc. 10. From Arts. 110 and 111, shew that approximately iQf log ft = log a -f log sin B - log sin A + -^-7 (cot A - cot B). 11. By continuing the approximation in Art. 106 so as to include the terms involving r 4 , shew that approximately py (a 2 - 3/3 2 - - - 12. From the preceding result shew that if A = A' + then approximately pysinA' ~ X. GEODETICAL OPERATIONS. 114. One of the most important applications of Trigono- metry, both Plane and Spherical, is to the determination of the figure and dimensions of the Earth itself, and of any portion of its surface. We shall give a brief outline of the subject, and for further information refer to Woodhouse's Trigonometry, to the article Geodesy in the English Cyclopcedia, and to Airy's treatise on the Figure of the Earth in the Encyclopedia Metropolitans For practical knowledge of the details of the operations it will be necessary to study some of the published accounts of the great surveys which have been effected in different parts of the world, as for example, the Account of the measurement of two sections of the Meridional arc of India, by Lieut.-Colonel Everest, 1847 ; or the Account of the Observations and Calculations of the Prin- cipal Triangulation in the Ordnance Survey of Great Britain and Ireland, 1858. 115. An important part of any survey consists in the mea- surement of a horizontal line, which is called a base. A level plain of a few miles in length is selected and a line is measured on it with every precaution to ensure accuracy. Hods of deal, and of metal, hollow tubes of glass, and steel chains, have been used in different surveys ; the temperature is carefully observed during the opera- tions, and allowance is made for the varying lengths of the rods or chains, which arise from variations in the temperature. 116. At various points of the country suitable stations are selected and signals erected ; then by supposing lines to be drawn connecting the signals, the country is divided into a series of triangles. The angles of these triangles are observed, that is, the angles which any two signals subtend at a third. For example, suppose A and B to denote the extremities of the base, and C a 86 GEODETICAL OPERATIONS. signal at a third point visible from A and B ; then in the triangle AEG the angles ABC and BAG are observed, and then AC and BC can be calculated. Again, let D be a signal at a fourth point, such that it is visible from C and A ; then the angles ACD and (2MZ) are observed, and as AC is known, CD and -4Z) can be calculated. 117. Besides the original base other lines are measured in convenient parts of the country surveyed, and their measured lengths are compared with their lengths obtained by calculation through a series of triangles from the original base. The degree of close- ness with which the measured length agrees with the calculated length is a test of the accuracy of the survey. During the pro- gress of the Ordnance Survey of Great Britain and Ireland, seve- ral lines have been measured ; the last two are, one near Lough Foyle in Ireland, which was measured in 1827 and 1828, and one on Salisbury Plain, which was measured in 1849. The line near Lough Foyle is nearly 8 miles long, and the line on Salisbury Plain is nearly 7 miles long ; and the difference between the length of the line on Salisbury Plain as measured and as calculated from the Lough Foyle base is less than 5 inches (An Account of the Observations. . . page 419). 118. There are different methods of effecting the calculations for determining the lengths of the sides of all the triangles in the survey. One method is to use the exact formulae of Spherical Trigonometry. The radius of the Earth may be considered known very approximately ; let this radius be denoted by r, then if a be the length of any arc the circular measure of the angle which the arc subtends at the centre of the earth is - . The formulse of f Spherical Trigonometry give expressions for the trigonometrical functions of - . so that - may be found and then a. Since in r > r J practice - is always very small, it becomes necessary to pay GEODETICAL OPERATIONS. 87 attention to the methods of securing accuracy in calculations which involve the logarithmic trigonometrical functions of small angles (Plane Trigonometry, Art. 205). Instead of the exact calculation of the triangles by Spherical Trigonometry, various methods of approximation have been pro- posed ; only two of these methods however have been much used. One method of approximation consists in deducing from the angles of the spherical triangles the angles of the chordal triangles, and then computing the latter triangles by Plane Trigonometry (see Art. 105). The other method of approximation consists in the use of Legendre's Theorem (see Art. 106). 119. The three methods which we have indicated were all used by Delambre in calculating the triangles in the French survey (Base du Systeme Metrique, Tome in. page 7). In the earlier operations of the Trigonometrical survey of Great Britain and Ireland, the triangles were calculated by the chord method ; but this has been for many years discontinued, and in place of it Legendre's Theorem has been universally adopted (An Account of the Observations ... page 244). The triangles in the Indian Survey are stated by Lieut. -Colonel Everest to be computed on Legendre's Theorem. (An Account of the Measurement ... page CLVIII.) 120. If the three angles of a plane triangle be observed, the fact that their sum ought to be equal to two right angles affords a test of the accuracy with which the observations are made. We shall proceed to shew how a test of the accuracy of observations of the angles of a spherical triangle formed on the Earth's surface may be obtained by means of the spherical excess. 121. The area of a spherical triangle formed on the Earths surface being known in square feet, it is required to establish a rule for computing the spherical excess in seconds. Let n be the number of seconds in the spherical excess, s the number of square feet in the area of the triangle, r the number of 88 GEODETICAL OPERATIONS. feet in the radius of the Earth. Then if E be the circular mea- sure of the spherical excess, and therefore 180.60.60 206265 nr* 5 "206265 ' Now by actual measurement the mean length of a degree on the Earth's surface is found to be 365155 feet; thus ^=365155. With the value of r obtained from this equation it is found by logarithmic calculation, that log n = log s -9-326774. Hence n is known when s is known. This formula is called General Roy's rule, as it was used by him in the Trigonometrical survey of Great Britain and Ireland. Mr Davies, however, claims it for Mr Dalby. (See Hutton's Course of Mathematics } by Davies, Vol. n. p. 47.) 122. In order to apply General Hoy's rule, we must know the area of the spherical triangle. Now the area is not known exactly unless the elements of the spherical triangle are known exactly ; but it is found that in such cases as occur in practice an approximate value of the area is sufficient. Suppose, for example, that we use the area of the plane triangle considered in Legendre's Theorem, instead of the area of the Spherical Triangle itself; then it appears from Art. 109, that the error is approximately denoted by the fraction - o*"" of the former area, and this fraction is less than *0001, if the sides do not exceed 100 miles in length. Or again, suppose we want to estimate the influence of errors in the angles on the calculation of the area; let the GEODETICAL OPERATIONS. 89 a/3 sin C circular measure of an error be h, so that instead of A asn i ,, ,, , ,, we ought to use - ' ; the error then bears to the area A approximately the ratio expressed by hcotC. Now in modern observations k will not exceed the circular measure of a few seconds, so that, if C be not very small, h cot C is practically in- sensible. 123. The following example was selected by Woodhouse from the triangles of the English survey, and has been adopted by other writers. The observed angles of a triangle being respectively 42. 2'. 32", 67. 55'. 39", 70. 1'. 48", the sum of the errors made in the observations is required, supposing the side opposite to the angle A to be 27404-2 feet. The area is calculated from the ex- a 2 sin B sin C , , ~ pression ^. - - , and by General Roys rule it is found Zi sin jA. that n=-23. Now the sum of the observed angles is 180-1", and as it ought to have been 180+ *23", it follows that the sum of the errors of the observations is 1"'23. This total error may be distributed among the observed angles in such proportion as the opinion of the observer may suggest ; one way is to increase each of the observed angles by one-third of 1"*23, and take the angles thus corrected for the true angles. 124. An investigation has been made with respect to the form of a triangle, in which errors in the observations of the angles will exercise the least influence on the lengths of the sides, and although the reasoning is allowed to be vague it may be deserving of the attention of the student. Suppose the three angles of a triangle observed, and one side, as a, known, it is required to find the form of the triangle in order that the other sides may be least affected by errors in the observations. The spherical excess of the triangle may be supposed known with sufficient accuracy for practice, and if the sum of the observed angles does not exceed two right angles by the proper spherical excess, let these angles be altered by adding the same quantity to 90 GEODETICAL OPERATIONS. each, so as to make their sum -correct. Let A, 5, (7 be the angles thus furnished by observation and altered if necessary ; and let 8.4, 85 and 8(7 denote the respective errors of A, B and (7. Then A + 85 + 8(7 = 0, because by supposition the sum of A, B and C is correct. Considering the triangle as approximately plane, the . a sin (C + 8(7) ., a sin (G + 8(7) true value of the side c is - -^ ^~- , that is, -; rr^-zrs - ft) (2). It will be observed that the angular co-ordinates here used are analogous to the latitude and longitude which serve to determine the positions of places on the Earth's surface ; is the complement of the latitude and is the longitude. 134. Equation (1) of the preceding Article may be written / 2 2 0\ thus : cos r { cos = + sin 2 - ) \ A AJ / ft f)\ i ( cos 2 7: sin 2 - ) + 2 sir \ J J/ = cos a ( cos -5 sin - I + z sin a sin - cos ^ cos f < p ). A ZJ J J a Divide by cos 2 ^ and rearrange ; hence f\ /i tan 2 ^ (cos r + cos a) - 2 tan - sin a cos ( ft) + cos r cos a = 0. GO Let tan - l and tan - denote the values of tan - found from 2S - this quadratic equation ; then by Algebra, Chapter xxn. 0. . cos r cos a a + r a r tan -^ tan -^ = = tan 75 tan . 2 2 cosr-fcosa 2 2 Thus the value of the product tan -^ tan -^ is independent of <; this result corresponds to the well-known property of a circle in Plane Geometry which is demonstrated in Euclid in. 36 Corollary. 135. Let three arcs OA, OB, 00 meet at a point. From any point P in OB draw PM perpendicular to OA, and PN perpen- dicular to 00. The student can easily draw the required diagram. IN PLANE AND SPHERICAL TRIGONOMETRY. 99 Then, by Art. 65, sinPJ/"=sin0Psin^OP, sin PN= sin OP sin COB ; sinPJf sin A OB therefore sin COB' Thus the ratio of sin PJ/ to sin P^ is independent of the posi- tion of P on the arc OB. 136. Conversely suppose that from any other point p arcs pm and pn are drawn perpendicular to OA and 00 respectively; then if sinprn __ sin PM it will follow that p is on the same great circle as and P. 137. From two points P l and P 2 arcs are drawn perpendi- cular to a fixed arc ; and from a point P on the same great circle as P l and P 2 a perpendicular is drawn to the same fixed arc. Let PPj = O l and PP 2 = 2 ; and let the perpendiculars drawn from P, P 1? and P 2 be denoted by x, x l9 and x 2 . Then will sin sin 0. sin x = - //a ' x sin a;. -f -rr^ 4n sin a,- . sin O + sin # + 2 Let the arc PjP,, produced if necessary, cut the fixed arc at a point ; let a denote the angle between the arcs. We will sup- pose that P l is between and P^ and that P is between 7J and P 2 . Then, by Art. 65, sin x l = sin a sin OP l = sin a sin (OP - 0^ = sin a (sin OP cos X - cos OP sin 0J ; sin x a = sin a sin OP 2 = sin a sin (OP 4- 2 ) = sin a (sin OP cos O s + cos OP sin 2 ). Multiply the former by sin 2) and the latter by sin O l9 and add ; thus sin O a sin x l -f sin t sin x a = sin (0 X + 2 ) sin a sin OP = sin (O l + O a ) sin x. H2 100 CONNEXION OF FORMULAE The student should convince himself by examination that the result holds for all relative positions of P, P l9 and P 2 , when due regard is paid to algebraical signs. 138. The principal use of Art. 137 is to determine whether three given points are on the same great circle; an illustration will be given in Art. 146. 139. The arcs drawn from the angles of a spherical triangle perpendicular to tlie opposite sides respectively meet at a point. Let CF be perpendicular to AB. From F suppose arcs drawn perpendicular to CB and CA respectively ; denote the former by and the latter by 77. Then, by Art. 135, _ sin rj sin FCA ' Eut, by Art. 65, cos B = cos CF sin FOB, cos A = cos CF sin FCA ; sin f cos B cos B cos C therefore - = - 7 = - -. - ~ . sin 17 cos A cos A cos C And if from any point in CF arcs are drawn perpendicular to CB and CA respectively, the ratio of the sine of the former perpen- dicular to the sine of the latter perpendicular is equal to - smr; by Art. 135. IN PLANE A&D SPHERICAL TRIGONOMETRY. 101 In like manner suppose AD perpendicular to EC ; then if from any point in AD arcs are drawn perpendicular to AC and AE respectively, the ratio of the sine of the former perpendicular to the sine of the latter perpendicular is equal to - ^ . cos A cos E Let CF and AD meet at P, and from P let perpendiculars be drawn on the sides a, b, c of the triangle ; and denote these per- pendiculars by x 9 y, z respectively : then we have shewn that sin x cos B cos G sin y cos A cos G ' , , . sin y cos A cos C and that -r - = -: = : sin z cosyicosj5 hence it follows that sin x _ cos B cos C sin z cos E cos A ' and this shews that the point P is on the arc drawn from E per- pendicular to AC. Thus the three perpendiculars meet at a point, and this point is determined by the relations sin x sin y sin z cos B cos G cos G cos A cos A cos 2? * 140. In the same manner it may be shewn that the arcs drawn from the angles of a spherical triangle to the middle points of the opposite sides meet at a point ; and if from this point arcs x, y, z are drawn perpendicular to the sides a, b, c respectively, sin x sin y sin z sin E sin G sin G sin A sin A sin E * 141. It is known in Plane Geometry that a certain circle touches the inscribed and escribed circles of any triangle ; this circle is called the Nine points circle : see Appendix to Euclid, pages 317, 318, and Plane Trigonometry, Chapter xxiv. 102 CONNEXION OF FORMULA We shall now shew that a small circle can always be deter- mined on the sphere to touch the inscribed and escribed circles of any spherical triangle. 142. Let a denote the distance from A of the pole of the small circle inscribed within a spherical triangle ABC. Suppose that a small circle of angular radius p touches this inscribed circle internally ; let /5 be the distance from A of the pole of this touch- ing circle ; let y be the angle between arcs drawn from A to the pole of the inscribed circle and the pole of the touching circle respectively. Then we must have cos (p r) = cos a cos /? + sin a sin f3 cos y (1). Suppose that this touching circle also touches externally the escribed circle of angular radius r^ ; then if c^ denote the distance from A of the pole of this escribed circle, we must have cos (p -f r^ = cos ttj cos P + sin a A sin ft cos y (2 ). Similarly, if a s and a 3 denote the distances from A of the poles of the other escribed circles, in order that the touching circle may touch these escribed circles externally, we must also have cos(p-f r 2 ) = cos a 2 cos /3 -f sina s sin/?cosf--yj (3), cos (p + r s ) = cos a 3 cos (3 + sin a 3 sin /5 cos Q + yj (4). We shall shew that real values of p, /?, and y can be found to satisfy these four equations. Eliminate cos y from (1) and (2) ; thus cos p (cos r sin a t cos r l sin a) + sin p (sin r sin a t + sin r l sin a) = cos )3 (cos a sin dj cos a x sin a) (5). Suppose that the inscribed circle touches AB at the distance m from A, and that the escribed circle of angular radius r l touches AB at the distance m l from A. Then, by Art. 65, IN PLANE AND SPHERICAL TRIGONOMETRY. 103 cot a = cot m cos , cos a = cos r cos ra, sin r = sin a sin -^ ,, . cosr cot a 1 A therefore = = . cos . sin a cos m sin m 2 Similarly we may connect a A and r : with m^ Thus we obtain from (5) ' Sm 2 A/ I 1 \ _ . COS p COS ( -; : 4- 2 Sin p ! 2 \sin7?i sin m / = cos fi cos (cot m cot raj ; therefore cos p (sin m^ - sin m) -f 2 sin p sin m sin w x tan ^ = cos )3 sin (rn> l ni). But by Arts. 89 and 90 we have m = s - a, and m l = 5 ; there- fore by the aid of Art. 45 we obtain 2 cos p sin - cos + 2n sin p = cos ft sin a (6), where n has the meaning assigned in Art. 46. In like manner if we eliminate siny between (3) and (4), putting m 2 for s c, and m 3 for s b, we obtain cos p (sin m a + sin w 3 ) - 2 sin p sin m a sin m a cot ^ =* cos /3 sin (m 2 4- ra 3 ), therefore 2 cos p sin cos ^ 2wsin p = cos ft sin a (7). From (6) and (7) we get . a . b . c sin ?rSin sm- 22 21 tanp= rc =^ tan ^ kv Art. 92 (8), b c cos - cos ^ cos p and cos ft = ~* (9). cos- 104< CONNEXION OF FORMULA We may suppose that cos -= is not less than cos or cos ~ , so -j 2t 2t that we are sure of a possible value of cos /3 from (9). It remains to shew that when p and ft are thus determined, all the four fundamental equations are satisfied. It will be observed that, p and fi being considered known, cosy can be found from (1) or (2), and 'siny can be found from (3) or (4) : we must therefore shew that (1) and (2) give the same value for cos y, and that (3) and (4) give the same value for sin y ; and we must also shew that these values satisfy the condition cos 2 y + sin 2 y - 1. From (1) we have cos p sin r ( cos 6\ . _ - ( cot r 4- tan p cos m cot r -- - ) = sm p cos y, sma \ cosp/ that is, . A ( , . . b c} cos p sin I cos (s -a) sins cos ^ cos ~ isin s + sin i a sin i 6 sin i c n a I cos- = sin /3 cos y ; this reduces to sin (6 + c) cos cos and it will be found that (2) reduces to the same ; so that (1) and (2) give the same value for cos y. In like manner it will be found that (3) and (4) agree in reducing to A ( . , ^ b c} cos p cos - . sm (c - b) cos ^ cos ^ 2 ) a . c o A - n - ) COS 2 Sm ^ --- : - a 2 cos .w IN PLANE AND SPHERICAL TRIGONOMETRY. 105 It only remains to shew that the condition cos 2 y 4- sin a y 1 is satisfied. COS/3 put X for cot r {1 Jc cos (s a)}, and Y for cot r t {1 & cos $}. Then (1) and (2) may be written respectively thus : ^ (X cos p + sin p) sin - = sin/? cosy .......... (10), (F cos p sin p) sin = sin /3 cos y . . ....... (11). From (10) and (11) by addition A (X + Y) sin cos p = 2 sin /3 cos y ; therefore 4sin 2 cos 2 y = (X 2 + F 2 + 2XY) sin 2 ^ cos 2 p... (12). 2s But from (10) and (11) by subtraction (X- Y) cos p = 2 sin p ; therefore (X 2 + F 2 ) cos 2 p - 4 sin 2 p + 2ZY cos 2 p. Substitute in (12) and we obtain sin 2 p cos 2 y = (sin 2 /a + XY cos 2 p) sin 2 -^ ....... (13). J Again, put JTj for cot r a {l & cos (5 c)}, and Y l for cot r 3 {l Jc cos (*' 6)}. Then (3) and (4) may be written respectively thus : A (X l cos p sin p) cos -^ = sin fi sin y ......... (14), (F x cos p - sin p) cos = sin f$ sin y ...... (15). From (14) and (15) by subtraction A A l - F x ) cos cos p = 2 sin /3 sin y, 106 CONNEXION OF FORMULAE and from (14) and (15) by addition, (X l + Y^ cos p = 2 sin p, whence sin 8 ^sin*y=(sin 2 p-jr i r i cosV)cos 2 - (16). Hence from (13) and (16) it follows that we have to establish the relation sin 2 p = sin 2 p + (XT sin 2 ^-XJ l cos 2 ^\ cos 2 p. But sin* /? = 1 - cos 2 /3 = sin 2 p + cos 2 p - 1c* cos 2 p, so that the re- lation reduces to Now Y V . ^A cotrcotr {! & cos $}{! & cos(s a)} sin(s 5)sin(sr) a 2 sin" = : - p A sin b sin c _ {1 &cos $}{!-& cos (s- a)} sin b sin c Similarly JT^cos 2 ^ J 2 sin 6 sin c Subtract the latter from the former ; then we obtain (s-b) + cos (s-c) - cos s -cos (-)} sin 5 sine COS 5 C S " "" C S "" C S "" ; cos - ,i,. that is . . sin 6 sin c 6-c 6+c) --- cos -^r- V 2 2 J cos &* ( 6 + c JIE^ ( cos rtf + c a cos ~~ - cos -- cos , . 5 . c 6 c 4 sin - sin - cos - cos ^ ^f 2222 ^ 2 r. a c-5 . f c + 6l that is . . + . . {sm 3 -^ -- sin 2 -^r >, sm 6 sin c sin 6 sin c ( 2 2 J that is 1 k* ; which was to be shewn. IN PLANE AND SPHERICAL TRIGONOMETRY. 107 143. Thus the existence of a circle which touches the in- scribed and escribed circles of any spherical triangle has been established. The distance of the pole of this touching circle from the angles B and G of the triangle will of course be determined by formulae corresponding to (9) ; and thus it follows that a c a b cos - cos 5 cos p cos ^ cos - cos p 2t 2i A J , and , o c COS - COS 2 must both be less than unity. 144. Since the circle which has been determined touches the inscribed circle internally and touches the escribed circles externally, it is obvious that it must meet all the sides of the spherical triangle. We will now determine the position of the points of meeting. Suppose the touching circle intersects the side AB at points distant X and p respectively from A. Then by Art. 134 we have a be . ^ cos n - cos s cos H X /* _ cos p cos p 2 22 .-. 2 2 ~~ cos p + cos 3 ~ a b c V ^ cos y -f cos ^ cos - In the same way we must have by symmetry b a c . cos ^ - cos ^ cos H c-X. c-u. 2 22 tan -5- tan -~-= 7 ....(2). 2 2 o a c cos y + cos ^ cos - From (2), when we substitute the value of tan ^ tan ^ given by (1), we obtain cos 2 -^ cos 2 ^ cos 2 - + cos 2 - sin 2 - tan - + tan | = b q 1 i cos ^ sin s (cos K + cos ^ cos ^} 108 CONNEXION OF FORMULA a be b . c cos ^ cos cos - cos ^ sin ^ 6 c' cos -sn- cos ^+ cos -cos - From (1) and (3) we see that we may put cos|-cos|cos| 2 = - b . c .................. (4) ' Cos 2 sm 2 5 . c cos^sin^ (5). 2 a b c cos + cos - cos J 2.2 Similar formulae of course hold for the points of intersection of the touching circle with the other sides. 145. Let z denote the perpendicular from the pole of the touching circle on AB\ then (A \ sin z = sin p sm ( - + y J of- A A ' = sm p [ sin -= cos y -f cos - \ 2 A . \ J-g-siny). But from (2) and (3) of Art. 142 we have A cospsin-^ , . ._ Zi / . Cb . C\ sm p cos y = ( Z sm sin -^ sin - ) , fi \ J ^a 2iJ where Z sin (s a) cos 5 sin (s - a) cos ^ cos ^ sec ^ , ^ cospcos- a 5 c \ and sin 8 sin y = ^(Z sin ^ sin - sin - ) , n \ * A & ZJ where Z^ - sin (s - b) cos (s - c) sin (5 - 6) cos ^ cos - sec - . IN PLANE AND SPHERICAL TRIGONOMETRY. 109 Therefore . 2 A _ 2 A . a . b . c sin 3 = 2 - sm (s a) sm (5 b) sm (s - c) ( - b c a) : ' . ; . 2 1 \ 1 - cos s cos ^ cos T . sec - } , sin b sine ( 222) and cos 2 - sm s sm (5 - a) sin (s b) (., , N b c a) = v . 7 ; - ' < 1 - cos (s - c) cos TT cos ~ sec K > . sin b sine (, 2 2 2J -4 A Therefore Z sin 2 -^ + Z^ cos 2 ^ is equal to the product of sin (5 a) sin (s 6) sin 6 sine into b c a ( . , x / x 1 sin v s - c) -f.sins cos ^ cos ^ sec < sin (s c) cos s + cos (5 c) sin s > sin (5 a) sin (s b) (~ . a + b c b c a . ^ = - . ( . * ^2sin 5- cos --cos^r cos - sec^r sin (2s - c) V sin 6 sm c ( 2 2 222 x j sin (s a) sin (5 b) C . a + b . . , v 5 2 sin 6 sin sin (5 a) sin (s - b) sin ^ cos ^ cos - I c a i sin 6 sin cos sin (5 a) sin (s 6) sin 2 - sin^ sin (5 a) sin (s b) sin 2 c a a 5 . c sm b sin %- cos - 2 cos cos - sm - A A a A A 110 CONNEXION OF FORMULAE Therefore sin T 2 sin 2 sui (s - a) sm (s b) cosp . a . b . c 2 v /v ' in ^ = - sin - - sin - sm - ^ ____-_-_-_. _ i w 222 c . sin - sm a sm o L ^ J cosp . a . b . c ( ' ^4 -7? _) ... = sui 2 sm - sm - < 2 cos 2 ; 1 v ; by (2) of Art. 54. cosp . a . b . c . . _ v Thus sinz = - sm - sm - sin - cos (A B) = sin p cos (A B). Similar expressions hold for the perpendiculars from the pole of the touching circle on the other sides of the spherical triangle. 146. Let P denote the point determined in Art. 139 ; G the point determined in Art. 140, and W the pole of the touching circle. We shall now shew that JP, G, and N are on a great circle. Let x, y, z denote the perpendiculars from N on the sides a, b, c respectively of the spherical triangle ; let x^ y^ z l denote the perpendiculars from P ' ; and x 2 , y^ z t the perpendiculars from G. Then by Arts. 145, 139, and 140 we have sin x sin y sin z cos (B - C) cos (G - -4) cos (A-B)' sin x l sin y l sin z l cos B cos G cos G cos A cos A cos B ' sin x^ sin y 9 sin z 2 sin B sin G ~ sin G sin A ~~ sin A sin B ' Hence it follows that sin x = t l sin x l + t 2 sin # 2 , sin # = ^ sin y x -K 2 sin y 2 , sin z = sin z + 1 sin z IN PLANE AND SPHERICAL TRIGONOMETRY. Ill where t l and t 2 are certain quantities the values of which are not required for our purpose. Therefore by Art. 137 a certain point in the same great circle as P and G is at the perpendicular distances x, y, z from the sides a, 5, c respectively of the spherical triangle : and hence this point must be the point W. 147. The resemblance of the results which have been obtained to those which are known respecting the Nine points circle in Plane Geometry will be easily seen. The result tanp^tanJS corresponds to the fact that the 2i radius of the Nine points circle is half the radius of the circum- scribing circle of the triangle. From equation (4) of Art. 144 by supposing the radius of the sphere to become infinite we obtain X = : this corresponds JLC to the fact that the Nine points circle passes through the feet of the perpendiculars from the angles of a triangle on the opposite sides. From equation (5) of Art. 144 by supposing the radius of the sphere to become infinite we obtain //, = - : this corresponds to the . Li fact that the Nine points circle passes through the middle points of the sides of a triangle. From Art. 145 by supposing the radius of the sphere to be- come infinite we obtain z = ^ R cos ( A B) : this is a known property of the Nine points circle. In Plane Geometry the points which correspond to the P, G, and N of Art. 146 are on a straight line. 148. The results which have been demonstrated with respect to the circle which touches the inscribed and escribed circles of a spherical triangle are mainly due to Dr Hart and Dr Salmon. See the Quarterly Journal of Mathematics, Vol. vi. page 67. 112 EXAMPLES. EXAMPLES. . a /( cosScos(S-A)) . 1. From the formula sin- = / -j - . . n . ' - > deduce 2 v I sin 2? sin (7 J . the expression for the area of a plane triangle, namely a 2 sin B sin C , ,, ,. . ,. . ^ : -j , when the radius of the sphere is indefinitely in- creased. 2. Two triangles ABC, abc, spherical or plane, equal in all respects, differ slightly in position : shew that cos ABb cos BCc cos CAa + cos ACc cos GBb cos BAa = 0. 3. Deduce formulae in Plane Trigonometry from Napier's Analogies. 4. Deduce formulae in Plane Trigonometry from Delambre's Analogies. K T A*- jf c A+ B . C , , 5. From the formula cos -cos = sin -cos ^- deduce 2i 2i A 2i the area of a plane triangle in terms of the sides and one of the angles. 6. What result is obtained from Example 7 to Chapter "VI., by supposing the radius of the sphere infinite 1 7. From the angle C of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the sides a and b: shew that this perpendicular makes an angle S B with the side a, and an angle S A with the side b. 8. From each angle of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the adjacent sides. Shew that these perpendiculars meet at a point; and that EXAMPLES. 113 if x, y, z are the perpendiculars from this point on the sides a, 6, c respectively, sin a; _ siny ' sins sin (S-) sin (S-C) ~~ sin (S-Cjam (S-A) = slrT(tf^l)~sin (S-) ' 9. Through each angle of a spherical triangle an arc is drawn so as to make the same angle with one side which the perpen- dicular on the base makes with the other side. Shew that these arcs meet at a point; and that if x, y, z are the perpendiculars from this point on the sides a, b, c respectively, sin x sin y sin z cos A cos B cos G ' 10. Shew that the points determined in Examples 8 and 9, and the point N of Art. 146 are on a great circle. State the corresponding theorem in Plane Geometry. 11. If one angle of a spherical triangle remains constant while the adjacent sides are increased, shew that the area and the sum of the angles are increased. 12. If the arcs bisecting two angles of a spherical triangle and terminated at the opposite sides are equal, the bisected angles will be equal provided their sum be less than 180. [Let BOD and COE denote these two arcs which are given equal. If the angles B and G are not equal suppose B the greater. Then CD is greater than BE by Art. 58. And as the angle OBG is greater than the angle 0GB, therefore 00 is greater than OB ; therefore OD is greater than OE. Hence the angle ODG is greater than the angle OEB, by Example 11. Then construct a spherical triangle BGF on the other side of BC, equal to GBE. Since the angle ODG is greater than the angle OEB, the angle FD-G is greater than the angle DFCy therefore CD is less than GF, so that GD is less than BE. See the corresponding problem in Plane Geometry in the Appendix to Euclid, page 317.] T. S. T. I XIII. POLYHEDBONS. 149. A polyhedron is a solid bounded by any number of plane rectilineal figures which are called its faces. A polyhedron is said to be regular when its faces are similar and equal regular polygons, and its solid angles equal to one another. 150. If $ be tJie number of solid angles in any polyhedron, F the number of its faces, E the number of its edges, then Take any point within the polyhedron as centre, and describe a sphere of radius r, and draw straight lines from the centre to each of the angular points of the polyhedron; let the points at which these straight lines meet the surface of the sphere be joined by arcs of great circles, so that the surface of the sphere is divided into as many polygons as the polyhedron has faces. Let s denote the sum of the angles of any one of these poly- gons, m the number of its sides ; then the area of the polygon is ^{s (m 2)?r} by Art. 99. The sum of the areas of all the polygons is the surface of the sphere, that is, irr 2 . Hence since the number of the polygons is F, we obtain Now 2$ denotes the sum of all the angles of the polygons, and is therefore equal to 2?r x the number of solid angles, that is, to '2irSy and %m is equal to the number of all the sides of all the polygons, that is, to 2E, since every edge gives rise to an arc which is common to two polygons. Therefore therefore S+F=E+2. POLYHEDRONS. 115 151. Tliere can be only five regular polyhedrons. Let m be the number of sides in each face of a regular poly- hedron, n the number of plane angles in each solid angle ; then the entire number of plane angles is expressed by mF, or by nS, or by 2E 3 thus from these equations we obtain ~_ 4m 2mn ._ 4% 2 (m + n) mn 9 2 (m -f n) mn 9 2 (m + T&) mn " These expressions must be positive integers, we must therefore have 2 (m + n) greater than mn ; therefore h - must be greater than ^ : m n 2 but T& cannot be less than 3, so that - cannot be greater than ~ , n o and therefore must be greater than ; and as m must be an m 6 integer and cannot be less than 3, the only admissible values of m are 3, 4, 5. It will be found on trial that the only values of m and n which satisfy all the necessary conditions are the following : each regular polyhedron derives its name from the number of its plane faces. 7/Zr n S E / Name of regular Polyhedron. 3 3 4 6 4 Tetrahedron or regular Pyramid. 4 3 8 12 6 Hexahedron or Cube. 3 4 6 12 8 Octahedron. 5 3 20 30 12 Dodecahedron. 3 5 12 30 20 Icosahedron. It will be seen that the demonstration establishes something more than the enunciation states ; for it is not assumed that the faces are equilateral and equiangular and all equal. It is in fact 12 116 POLYHEDRONS. demonstrated that, there cannot be more than Jive solids each of which has all its faces with the same number of sides, and all its solid angles formed with the same number of plane angles. 152. The sum of all the plane angles which form the solid angles of any polyhedron is 2(S2)7r. For if m denote the number of sides in any face of the poly- hedron, the sum of the interior angles of that face is (m 2)?r by Euclid I. 32, Cor. 1. Hence the sum of all the interior angles of all the faces is 2 (m 2) TT, that is SWTT - 2^, that is 2 (E- F)v, that is 2 (S - 2) v. 153. To find the inclination of two adjacent faces of a regular polyhedron. Let AB be the edge common to the two adjacent faces, C and D the centres of the faces ; bisect AB at E, and join CE and DE-, CE and DE will be perpendicular to AB, and the angle CED is the angle of inclination of the two adjacent faces ; we shall denote it by 7. In the plane containing CE and DE draw CO and DO at right angles to CE and DE respectively, and meeting at 0; about as centre describe a sphere meeting OA 9 OC, OE at a, c, e respectively, so that cae forms a spherical triangle. Since AB is perpendicular to CE and DE, it is perpendicular to the plane CED, therefore the plane AOB which contains AB is perpendicular to the plane CED \ hence the angle cea of the spherical triangle is a right angle. Let m be the number of sides in each face of the polyhedron, n the number of the plane angles which form each solid POLYHEDRONS. 117 alible. Then the angle ace = ACE=~ = : and the angle cae 2m m is half one of the n equal angles formed on the sphere round a, that is. cae = = - . From the right-angled triangle cae 2n n cos cae = cos cOe sin ace, 7T flT I\ . TT that is cos - = cos ( ^ - - ) sin : n \'2 2/ //&' 7T cos- therefore sin - = . sin 154. ^o y&ttd ^6 rac^ o/" the inscribed and circumscribed spheres of a regular polyhedron. Let the edge AB = a, let OC = r and OA = fi, so that r is the radius of the inscribed sphere, and R is the radius of the circumscribed sphere. Then CE = AEwi AGE = ~ cot - , 2 m r = CE tan CEO = (7^ tan = cot - tan^; J J m ^ also r = It cos aOc = JR cot eca cot eac = ^ cot cot - ; 7tt 76 therefore 7t = r tan tan - = ^ tan - tan - . m n * 2 n 155. To find the surface and volume of a regular polyhedron. 2 The area of one face of the polyhedron is r- cot - , and therefore the surface of the polyhedron is . cot . r J 4: m Also the volume of the pyramid which has one face of the 2 polyhedron for base and for vertex is ^ . j- cot , and therefore the volume of the polyhedron is =-^ cot . 118 POLYHEDRONS. 156. To find the volume of a parallelepiped in terms of its edges and tlieir inclinations to one another. C Let the edges be OA = a, OB = b, OC = c; let the inclinations be BOO = a, CO A = (3,AOB = y. Draw CE perpendicular to the plane AOB meeting it at E. Describe a sphere with as a centre, meeting OA, OB, OC, OE at a, b, c, e respectively. The volume of the parallelepiped is equal to the product of its base and altitude = ab sin y . GE = abc sin y sin cOe. The spherical triangle cae is right-angled at e ; thus sin cOe = sin cOa sin cae = sin /? sin cab, and from the spherical triangle cab 7 J(I - cos 2 a - cos 2 8 - cos 2 y + 2 cos a cos 6 cos 7) sin cab = ^L> - d. __f - C - L' ; sin p sin y therefore the volume of the parallelepiped = abc */(! cos 2 a cos 2 /J cos 3 y + 2 cos a cos /? cos y). 157. To find the diagonal of a parallelepiped in terms of tlie three edges which it meets and their inclinations to one another. Let the edges be OA = a, 0-5 = 6, 0(7 = c; let the inclinations be BOC = a, COA=/3, AOB = y. Let OD be the diagonal re- quired, and let OE be the diagonal of the face OAB. Then = a* + b* + 2ab cos y + c 8 + ZcOE cos C'O.S'. POLYHEDRONS. 119 Describe a sphere with as centre meeting OA, OJB, OC, OE at #, 5, c, e respectively; then (see Example 14, Chap, iv.) a ~ cos cOb sin aOe + cos cOa sin 50e COS CUB = : 7-7 sin a Ob cos a sin a Oe + cos /3 sin bOe siny therefore OZ) 2 = a 2 + 5 2 + c 2 4- 2a6 cos y + ^ - (cos a sin a0e -f cos ft sin i and OE sin a6te = 5 sin y, 0^7 sin 6(9e = a sin y ; therefore OD 2 = ct 2 + 6 2 + c 2 + 2a6 cos y + 26c cos a + 2ca cos /?. 158. To find the volume of a tetrahedron. A tetrahedron is one-sixth of a parallelepiped which has the same altitude and its base double that of the tetrahedron ; thus if the edges and their inclinations are given we can take one-sixth of the expression for the volume in Art. 156. The volume of a tetrahedron may also be expressed in terms of its six edges ; for in the figure of Art. 156 let EG = a', CA = b f , AB = c'-, then - - - cos a = - try - , cos p = - 7r - - , cos y = - ^7 2bc 2ca '2ab and if these values are substituted for cos a, cos /?, and cos y in the expression obtained in Art. 156, the volume of the tetrahe- dron will be expressed in terms of its six edges. The following result will be obtained, in which F denotes the volume of the tetrahedron, 120 POLYHEDRONS. U4:V 2 --=-a' a b' 2 c' 2 + a 2 a' 2 (V 2 + c' 2 - a' 2 ) + b 2 b' 2 (c' 2 + a' 2 - b' 2 ) + c 2 c' 2 (a' 2 + b' 2 - c' 2 ) - a' 2 (a 2 - b 2 ) (a 2 - c 2 ) - b' 2 (b 2 - c 2 ) (b 2 - a 2 ) - c' 2 (c 2 - a 2 ) (c 2 - b 2 ). Thus for a regular tetrahedron we have 144 V 2 = 2a 6 . 159. If the vertex of a tetrahedron be supposed to be situ- ated at any point in the plane of its base, the volume vanishes ; hence if we equate to zero the expression on the right-hand side of the equation just given, we obtain a relation which must hold among the six straight lines which join four points taken arbi- trarily in a plane. Or we may adopt Carnot's method, in which this relation is established independently, and the expression for the volume of a tetrahedron is deduced from it ; this we shall now shew, and we shall add some other investigations which are also given by Carnot. It will be convenient to alter the notation hitherto used, by interchanging the accented and unaccented letters. 160. To find the relation holding among the six straight lines which join four points taken arbitrarily in a plane. Let A, By C, D be the four points. Let AB = c, C = a, CA = b; also let DA = a', DB = b', DC = c'. If D falls within the triangle ABC, the sum of the angles ADB, BDC, CD A is equal to four right angles ; so that cos ALB = cos (BDC + CD A). Hence by ordinary transformations we deduce 1 = cos 2 ADB + cos 2 BDC + cos 2 CD A - 2 cos ALB cos BDC cos CD A . If D falls witJiout the triangle ABC, one of the three angles at D is equal to the sum of the other two, and the result just given still holds. Now cos ADB = -fr-, , and the other cosines may be expressed in a similar manner; substitute these values in the POLYHEDRONS. 121 above result, and we obtain the required relation, which after reduction may be exhibited thus, 0=-a'6V + a' 2 a 2 (b 2 + c 2 - a 2 ) + b' 2 b 2 (c 2 + a 2 - b 2 ) + c'V (a 2 + b 2 - c 2 ) - a 2 (a' 2 - b' 2 ) (a' 2 - c' 2 ) - b 2 (b' 2 - c' 2 ) (b' 2 - a' 2 ) - c 2 (c' 2 - a' 2 ) (c' 2 - b' 2 ). 161. To express the volume of a tetrahedron in terms of its six edges. Let a, b, c be the lengths of the sides of a triangle ABC forming one face of the tetrahedron, which we may call its base ; let a', b', c' be the lengths of the straight lines which join A, B^ C respectively to the vertex of the tetrahedron. Let p be the length of the perpendicular from the vertex on the base ; then the lengths of the straight lines drawn from the foot of the perpendicular to A,JB,C respectively are J(a' 2 -p*) J(b' 2 -p 2 ), J(c' 2 -p 2 ). Hence the relation given in Art. 160 will hold if we put kj(a' 2 p 2 ) in- stead of a', ,J(b' 2 p 2 ) instead of b' } and fj(c 2 p 2 ) instead of c. We shall thus obtain p 2 (2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2 - a 4 - b 4 - c 4 ) - - aW + a' 2 a 2 (b 2 + c 2 - a 2 ) + b' 2 b* (c 2 + a 2 - b 2 ) + cV (a 2 + b 2 - c 2 ) - a 2 (a' 2 - b' 2 ) (a' 2 - c' 2 ) - b 2 (b' 2 - c' 2 ) (b' 2 - a' 2 ) - c 2 (c' 2 - a' 2 ) (c' 2 - b' 2 ). The coefficient of p 2 in this equation is sixteen times the square of the area of the triangle ABC ; so that the left-hand member is 144 F 2 , where F denotes the volume of the tetrahe- dron. Hence the required expression is obtained. 162. To find the relation holding among the six arcs of great circles which join four points taken arbitrarily on the surface of a sphere. Let A 9 By (7, D be the four points. Let AJB = y, BG = a, CA =/3', let DA =a', DB = f$ f , >C = y'. As in Art. 160 we have 1 = co$ 2 ADB + cos 2 J3DC + cos 2 CD A - 2 cos ADB cos BDC cos CD A. 122 POLYHEDRONS. XT A r> r> cos y cos a' cos ' jNow cos ADB= -- '-; . -,- 1 - and the other cosines sin a sin /5 may be expressed in a similar manner ; substitute these values in the above result, and we obtain the required relation, which after reduction may be exhibited thus, 1 = cos 2 a + cos 2 fi + cos 2 y 4- cos 2 a! + cos 2 ft + cos 2 y' cos 2 a cos 2 a' - cos 2 ft cos* /3' - cos 2 y cos 2 y' 2 (cos a cos ft cos y + cos a cos /3' cos y' 4- cos /3 cos a' cos y' + cos y cos a! cos /?') + 2 (cos a cos ^ cos a' cos /3' + cos ft cos y cos ft cos y' 4- cos y cos a cos y' cos a'). 163. To find the radius of the sphere circumscribing a tetra- fadron. Denote the edges of the tetrahedron as in Art. 161. Let the sphere be supposed to be circumscribed about the tetrahedron, and draw on the sphere the six arcs of great circles joining the angular points of the tetrahedron. Then the relation given in Art. 162 holds among the cosines of these six arcs. Let r denote the radius of the sphere. Then and the other cosines may be expressed in a similar manner. Substitute these values in the result of Art. 162, and we obtain, after reduction, with the aid of Art. 161, 4x144 FV = 2a*b*a'*b'* + 26 WV 2 -r 2cW V - a*a' 4 - b*b'* - cY 4 . The right-hand member may also be put into factors, as we see by recollecting the mode in which the expression for the area of a triangle is put into factors. Let oaf + W + cc' = 2? + n* + 1? + m? + n.* + . . . } = f, by Art. 165, o where S is the number of the solid angles of the regular poly- hedron. ON THE SURFACE OF A SPHEKE. 133 Thus the sum of the squares of the cosines of tJie arcs which join any point on the surface of the sphere to the solid angles of the regular polyhedron is one third of the number of the solid angles. 178. Since P=Q = R in the preceding Article, it will follow that when the fixed points of Art. 174 are the solid angles of a regular polyhedron, then for any position of the spherical tri- angle ABC we shall have p = 0, q 0, and r = 0. For taking any position for the spherical triangle ABC we have 2 = PX 2 + QJJ? + Rv* + 2 then at A we have //, = and v = 0, so that P is then the value of 2 ; similarly Q and R are the values of 2 at B and C respec- tively. But by Art. 177 we have the same 'value for 2 whatever be the position of T ', thus P = therefore = 2pp.v + 2qvX + 2rX/z. This holds then for every position of T. Suppose T is at any point of the great circle of which A is the pole ; then X = : thus we get ppv = ; and therefore p = 0. Similarly q = 0, and r = 0. 179. Let there be any number of fixed points on the surface of a sphere ; denote them by H l9 H^ ZT 3 , . . . ; from any two points T and U on the surface of the sphere arcs are drawn to the fixed points : it is required to find the sum of the products of the cor- responding cosines, that is cos TH l cos UH l + cos TH 2 cos UH 2 + cos TII 3 cos Vff B + ... Let the notation be the same as in Art. 174 ; and let X', /*', v be the cosines of the arcs which join U with A, B, C respectively. Then by Art. 166, cos TH l cos UH l = ( 134* ARCS DRAWN TO FIXED POINTS. Similar results hold for cos TH Z cos UH 2 , cos ?W a cos Uff a , . . . Hence, with the notation of Art. 174, the required sum is XX' P + w'Q 4- w'R + (fJiv + VIL)P + (v\' + \v) q + (\p! + /A') r. Now by properly choosing the position of the triangle ABC we have p, q, and r each zero as in Art. 174; and thus the required sum becomes XX'P + p.pfQ + w'R. 180. The result obtained in Art. 174 may be considered as a particular case of that just given; namely the case in which the points T and U coincide. 181. A sphere is described about a regular polyhedron ; from any two points on the surface of the sphere arcs are drawn to the solid angles of the polyhedron : it is required to find the sum of the products of the corresponding cosines. With the notation of Art. 179 we see that the sum is And here P =Q = R = -^, by Art. 177. Thus the sum = f (XX' + //// + w r ) = ^ cos TU. O i) Thus the sum of the products of the cosines is equal to the product of the cosine of TU into a third of the number of the solid angles of the regular polyhedron. 182. The result obtained in Art. 177 may be considered as a particular case of that just given; namely, the case in which the points T and U coincide. 183. If TU is a quadrant then cos TU is zero, and the sum of the products of the cosines in Art. 181 is zero. The results p - 0, q = 0, r = 0, are easily seen to be all special ex- amples of this particular case. ( 135 ) XV. MISCELLANEOUS PEOPOSITIONS. 184. To find the locus of the vertex of a spherical triangle of given base and area. Let AB be the given base, -c suppose, AC = Q, Since the area is given the spherical excess is known; denote it by E' } then by Art. 103, cot J E- cot \ cot \ c cosec 4- cot ; therefore sin ( - J E) = cot \ cot J c sin |- E ; therefore 2 cot J c sin J E cos 2 - = sin sin (< - 1 E) ; therefore cos ^ cot J c sin ^ E + sin cos ( - J j^ + ^ j = cot -| c sin ^ jE'. Comparing this with equation (1) of Art. 133, we see that the required locus is a circle. If we call a, ft the angular co-ordinates of its pole, we have _ cot Jcsi It may be presumed from symmetry that the pole of this circle is in the great circle which bisects AJ3 at right angles ; and this presumption is easily verified. For the equation to that great circle is cos < - T = cos cos ( - - j + sin sin ( ^ - - J and the values a, < = ft satisfy this equation. 136 MISCELLANEOUS PROPOSITIONS. 185. To find the angular distance between the poles of the inscribed and circumscribed circles of a triangle. Let P denote the pole of the inscribed circle, and Q the pole of the circumscribed circle of a triangle ABC ; then by Art, 89, and QAB = S-C, by Art. 92; hence and cos PQ - cos PA cos QA + sin PA sin QA cos | (B C). Now, by Art. 62 (see the figure of Art. 89), cos PA = cos PE cosAE = cos r cos (s - a), sin PE sin r sin PAE sin \A ' thus cos PQ = cos R cos r cos (s a) + sin 72 sin r cos J (7? 0) cosec J -4. Therefore, by Art. 54 cos PQ = cos 72 cos r cos (s - a) + sin 72 sin r sin J (6 + c) cosec J a, therefore ; = cot r cos (s - a) + tan R sin A (b + c) cosec i a. cos72smr v y 2V / , T sin * _ 2siniasin A b sinic Now cotr= , tan/t= , n n therefore n . = -! sin s cos (s-a) + 2 sin i (6+c) sin i& sin ic i cos ^ sin r w t * J = ^ (sin a -f sin 5 + sin c). / cosPQ \ 2 1 Hence ( ^ . I 1 = -75- (sin a + sin o + sin c) - 1 \cos Jt sin r/ 47i z x = (cot r + tan R)* ( b 7 Al ' t - 94 ) ; therefore cos 2 PQ = cos 2 R sin 2 r + cos 2 (R - r), and sin 2 PQ = sin 2 (R-r)- cos 2 72 sin 2 r. MISCELLANEOUS PKOPOSITIONS. 137 186. To find the angular distance between tlie pole of the circumscribed circle and the pole of one of the escribed circles of a triangle. Let Q denote the pole of the circumscribed circle, and Q l the pole of the escribed circle opposite to the angle A. Then it may be shewn that QQ 1 = | TT + \ (C - A), and cos QQ 1 = cos E cos r l cos (s c) sin R sin r l sin J (C A) sec | B = cos R cos r 1 cos (s c) sin R sin r l sin J (c a) cosec J b. Therefore - ^= cot r. cos (s c) tan R sin A (c - a) cosec A 6 : by reducing as in the preceding Article, the right-hand member of the last equation becomes 1 , . (sin b + sin c sin a) ; hence therefore cos 2 $(?! = cos 2 R sin 2 r, + cos 2 (R + rj, and sin 2 QQ l = sin 2 (^ + rj - cos 2 J? sin 2 r r 187. ^TAe arc which passes through the middle points of the sides of any triangle upon a given base will meet the base produced at a fixed point, the distance of which from the middle point of tlie base is a quadrant. Let ABC be" any triangle, E the middle point of AC, and F the middle point of AB ; let the arc which joins E and F when produced meet BC produced at Q. Then sin BQ _ sin BFQ smAQ _$mAFQ ., , therefore 138 MISCELLANEOUS PROPOSITIONS. sin CQ sinAQF similarly -, T = -. ^ ; sin ^40 smCQF* therefore sin BQ = sin CQ ; therefore BQ -f CQ = IT. Hence if D be the middle point of BG 188. If three arcs le drawn from the angles of a spherical triangle through any point to meet the opposite sides, the products of the sines of the alternate segments of the sides are equal. Let P be any point, and let arcs be drawn from the angles A, B, C passing through P and meeting the opposite sides at D, E, F. Then smBD smBPD sinCT) sinCTD ~sin.DP 9 sin(7P csi- therefore . ., . sin (7^ , Similar expressions may be found for -r-^ and smAE and hence it follows obviously that sin BD sin CE sin AF __ sin CD smAE sin BF~ ' therefore sin BD sin CE sinAF = smCD sinAE sin BF. MISCELLANEOUS PROPOSITIONS. 139 189. Conversely, when the points D, E, F in the sides of a spherical triangle are such that the relation given in the preceding Article holds, the arcs which join these points with the opposite angles respectively pass through a common point. Hence the following propositions may be established : the perpendiculars from the angles of a spherical triangle on the opposite sides meet at a point ; the arcs which bisect the angles of a spherical triangle meet at a point ; the arcs which join the angles of a spherical triangle with the middle points of the opposite sides meet at a point ; the arcs which join the angles of a spherical triangle with the points where the inscribed circle touches the opposite sides respectively meet at a point. Another mode of establishing such propositions has been exemplified in A_rts. 139 and 140. 190. If AB and A'B' be any two equal arcs, and the arcs AA' and BB' be bisected at right angles by arcs meeting at P, then AB and A'B' subtend equal angles at P. For PA = PA' and PB = PB' ; hence the sides of the triangle PAB are respectively equal to those of PA'B' ; therefore the angle APB = the angle A'PB'. This simple proposition has an important application to the motion of a rigid body of which one point is fixed. For conceive a sphere capable of motion round its centre which is fixed ; then it appears from this proposition that any two fixed points on the 140 MISCELLANEOUS PROPOSITIONS. sphere, as A and B, can be brought into any other positions, as A' and B', by rotation round an axis passing through the centre of the sphere and a certain point P. Hence it may be inferred that any change of position in a rigid body, of which one point is fixed, may be effected by rotation round some axis through the fixed point. (De Morgan's Differential and Integral Calculus, page 489.) 191. Let P denote any point within any plane angle AOB, and from P draw perpendiculars on the straight lines OA and OB ; then it is evident that these perpendiculars include an angle which is the supplement of the angle AOB. The corresponding fact with respect to a solid angle is worthy of notice. Let there be a solid angle formed by three plane angles, meeting at a point 0. From any point P within the solid angle, draw perpendicu- lars PL, PM, PN on the three planes which form the solid angle ; then the spherical triangle which corresponds to the three planes LPMj MPN, NPL is the polar triangle of the spherical triangle which corresponds to the solid angle at 0, This remark is due to Professor De Morgan. 192. Suppose three straight lines to meet at a point and form a solid angle ; let a, /?, and y denote the angles contained by these three straight lines taken in pairs : then it has been proposed to call the expression ^/(l cos 2 a - cos* ft cos 2 y + 2 cos a cos fi cos y), the sine of the solid angle. See Baltzer's Theorie...der Determi- nanten, 2nd edition, page 177. Adopting this definition it is easy to shew that the sine of a solid angle lies between zero and unity. We know that the area of a plane triangle is half the product of two sides into the sine of the included angle : by Art. 156 we have the following analogous proposition ; the volume of a tetra- hedron is one sixth of the product of three edges into the sine of the solid angle which they form. Again, we know in mechanics that if three forces acting at a point are in equilibrium, each force is as the sine of the angle between the directions of the other two : the following proposition is analogous ; if four forces acting at a point are in equilibrium MISCELLANEOUS PKOPOSITIONS. each force is as the sine of the solid angle formed by the directions of the other three. See Statics, Chapter II. 193. Let a sphere be described about a regular polyhedron ; let perpendiculars be drawn from the centre of the sphere on the faces of the polyhedron, and produced to meet the surface of the sphere : then it is obvious from symmetry that the points of intersection must be the angular points of another regular poly- hedron. This may be verified. It will be found on examination that if S be the number of solid angles, and F the number of faces of one regular polyhedron, then another regular polyhedron exists which has S faces and F solid angles. See Art. 151. 194. Polyhedrons. The result in Art. 150 was first obtained by Euler; the demonstration which is there given is due to Legendre. The demonstration shews that the result is true in many cases in which the polyhedron has re-entrant solid angles ; for all that is necessary for the demonstration is, that it shall be possible to take a point within the polyhedron as the centre of a sphere, so that the polygons, formed as in Art. 150, shall not have any coincident portions. The result, however, is generally true, even in cases in which the condition required by the demonstra- tion of Art. 150 is not satisfied. We shall accordingly give another demonstration, and shall then deduce some important consequences from the result. We begin with a theorem which is due to Cauchy. 195. Lei there "be any network of rectilineal figures, not neces- sarily in one plane, but not forming a closed surface ; let E be the number of edges, F the number of figures, and S the number of corner points : then F + S = E + 1. This theorem is obviously true in the case of a single plane figure; for then F=\, and S=E. It can be shewn to be gene- rally true by induction. For assume the theorem to be true for a network of F figures ; and suppose that a rectilineal figure of n sides is added to this network, so that the network and the additional figure have m sides coincident, and therefore m + 1 142 MISCELLANEOUS PROPOSITIONS. corner points coincident. And with respect to the new network which is thus formed, let E r , F', S' denote the same things as E, F, S with respect to the old network. Then E' = E + n-m, F' = F+l, S' = 8 + n-(m + 1); therefore F' + S' - E' = F + S - E. But F+ S= E + 1, by hypothesis ; therefore I" + S f = E' + 1. 196. To demonstrate Euler's theorem we suppose one face of a polyhedron removed, and we thus obtain a network of recti- lineal figures to which Cauchy's theorem is applicable. Thus therefore 197. In any polyhedron the number of faces with an odd number of sides is even, and the number of solid angles formed with an odd number of plane angles is even. Let a, by c, d, ...... denote respectively the numbers of faces which are triangles, quadrilaterals, pentagons, hexagons, ...... Let a, (3, y, 8, ...... denote respectively the numbers of the solid angles which are formed with three, four, five, six, ...... plane angles. Then, each edge belongs to two faces, and terminates at two solid angles ; therefore From these relations it follows that a + c + e + ...... , and a + y + c + ...... are even numbers. 198. "With the notation of the preceding Article we have From these combined with the former relations we obtain 2^-3^=6+ 2c+ Thus 2E cannot be less than 3F } or less than 3. MISCELLANEOUS PROPOSITIONS. 199. From the expressions for E, F, and 8, given in the two preceding Articles, combined with the result 2F + 2S = 4 -t- 212, we obtain 2(a + b + c + d+ ...) + 2(a + j8 2(a + b + c + d+...) + 2(a + p therefore 2(a + +y + S + ...)-( + 26 + 3c + 4d+ ...) = 4 ... (1), 2( a + 6 + c + d + Therefore, by addition ...... = 8. the number of triangular faces together with the number of solid angles formed with three plane angles cannot be kss than eight. Again, from (1) and (2), by eliminating a, we obtain 3a + 2b + c-e-2f-~ - 2/5 - 4y - ...... = 12, so that 3a + 2b + c cannot be less than 12. From this result various inferences can be drawn ; thus for example, a solid cannot be formed which shall have no triangular, quadrilateral, or pen- tagonal faces. In like manner, we can shew that 3 a + 2/3 + y cannot be less than 12. 200. Poinsot has shewn that in addition to the five well- known regular polyhedrons, four other solids exist which are perfectly symmetrical in shape, and which might therefore also be called regular. We may give an idea of the nature of Poinsot's results by referring to the case of a polygon. Suppose five points A, B, C, D, E, placed in succession at equal distances round the circumference of a circle. If we draw a straight line from each point to the next point, we form an ordinary regular pentagon. Suppose however we join the points by straight lines in the fol- lowing order, A to C, C to E, E to B, B to D, D to A ; we thv.s form a star-shaped symmetrical figure, which might be considered a regular pentagon. 144 MISCELLANEOUS PROPOSITIONS. It appears that, in a like manner, four, and only four, new regular solids can be formed. To such solids, the faces of which intersect and cross, Euler's theorem does not apply. 201. Let us return to Art. 195, and suppose e the number of edges in the bounding contour, and e f the number of edges within it ; also suppose s the number of corners in the bounding contour, and s the number within it. Then E = e + e r ; S = s + s' ; therefore 1 + e + e = s + s' + F. But e = s ; therefore 1 -f e = s + F. We can now demonstrate an extension of Euler's theorem, which has been given by Cauchy. 202. Let a polyhedron be decomposed into any number of polyhedrons at pleasure; let P be the number thus formed, S the number of solid angles , F the number of faces y E the number of edges: then S + F = E + P+1. For suppose all the polyhedrons united, by starting with one and adding one at a time. Let e,f, s be respectively the num- bers of edges, faces, and solid angles in the first ; let e', f', s' be respectively the numbers of edges, faces, and solid angles in the second which are not common to it and the first; let e",f", s" be respectively the numbers of edges, faces, and solid angles in the third which are not common to it and the first or second; and so on. Then we have the following results, namely, the first by Art. 196, and the others by Art. 201; By addition, since s + s r + s" 4- . . . = S, f+f +/' -f . . . = F, and e -f e' -f Q" + . .. = E, we obtain MISCELLANEOUS EXAMPLES. 145 203. The following references will be useful to those who study the theory of polyhedrons. Euler, Novi Commentarii Academice....Petropolitance, Yol. iv. 1758; Legendre, Geometrie ; Poinsot, Journal de VEcole Poly technique, Cahier x ; Cauchy, Journal de VEcole Poly technique, Cahier xvi; Poinsot and Bertrand, Comptes Rendus...de V Academic des Sciences, Vol. XLVI ; Catalan, Theoremes et Problemes de Geometrie Elementaire ; Kirkman, Phi- losophical Transactions for 1856 and subsequent years; Listing, Abhandlungen der Koniglichen Gesellschaft...zu Gottingen, Yol. x. MISCELLANEOUS EXAMPLES. 1. Find the locus of the vertices of all right-angled spherical triangles having the same hypotenuse; and from the equation obtained, prove that the locus is a circle when the radius of the sphere is infinite. 2. AB is an arc of a great circle on the surface of a sphere, its middle point : shew that the locus of the point P, such that the angle APG = the angle BPG, consists of two great circles at right angles to one another. Explain this when the triangle becomes plane. 3. On a given arc of a sphere, spherical triangles of equal area are described : shew that the locus of the angular point opposite to the given arc is defined by the equation * tan- 1 /*"*(* + *)} + tan- f ( sintf J I si sine/ _, tan + tan * { -.-, \ ,( tan<9 } yv } + tan * { -r. - TT > --= /?, + <) J (sin (a -