But if AD be without the triangle, BAD-CAD=BAC, and there fore sin (AB+AC): sin (AB-AC): : cot i BAD+CAD) : tan { BAC :: cot BAC : tan 1 (BAD+CAD), sin (AB+AC): sin (AB-AC) :: cot BAC : tan 1 (BAD+CAD). Wherefore, &c. Q. E. D. LEMMA. The sum of the tangents of any two arches, is to the difference of their tangents, as the sine of the sum of the arches, to the sine of their difference. cos A cos B Let A and B be two arches, tan A + tan B : tan Atan B : : 'sin (A+B): sin (A-B). For, by $ 6. page 243, sin A Xcos B +cos AX sin B=sin (A+B), sin A sin B and therefore dividing all by eos A cos B, + sin (A + B) sin A that is, because = tan A, tan Attan B cos A x cos B' cos A sin (A+B) In the same manner it is proved that tan A-tan B cos A Xcos B sin (A-B) Therefore tan A 7 tan B : tan A-tan B :: sin cos A Xcos B* (A+B): sin (A-B). Q. E. D. PROP. XXXI. The sine of half the sum of any two angles of a spherical triangle is to the sine of half their difference, as the tangent of half the side adjacent to these angles is to the tangent of half the difference of the sides opposite to them; and the cosine of half the sum of the same angles is to the cosine of half their difference, as the tangent of half the side adjacent to them, to the tangent of half the sum of the sides opposite. Let C+B=2S, C-B=2D, the base BC=2B, and the difference of the segments of the base, or BD-CD=2X. Then, because (30.) sin (C+B) : sin (C-B): : tan BC : tan} (BD-CD), sin 28: sin 2D :: tan 'B : tan x. Now, sin 2S=sin ($ + S)=2 sin S x cos' S, (Sect. III. cor. Pl. Tr.). In the same manner, sin 2D= 2 sin Dx cos D. Therefore sin S x cos S; sin D X eos D :: tan B : tan X. A B tan B D B D с Again, in the spherical triangle ABC it has been proved, that siz C+sin B : sin C-sin B :: sin AB+sin AC : sin AB -sin AC, and since sin C + sin B=2 sin (C+B) X cos (C-B), (Sect. III. 7. PL Tr.)= 2 sin $Xcos D; and sin C - sin B=2 cos (C+B) X sin (C-B) = 2 cos 8 X sin D. Therefore 2 sin Sxcos D : 2 cos S X sin D :: sin AB + sin AC : sin AB - sin AC. But (3. Pl. Tr. sin AB+sin AC :sin AB -sin AC :: tan} (AB+AC): tan (AB-AC):: tan : tan 4, s being equal to ! (AB + AC) and 4 to 1 (AB-AC.), Therefore sin $Xcos D : cos Sxsin D :: tan & : tan 4. Since then tan X sin D Xcos D tan A cos SXsin D and , by multiplying equals tan B sin SX cos s tan sin S Xcos D tan X tan A (sin D)? X cos SX cos D (siņ D)2 by equals, X tan Σ (sin 5)2 X cos S X cos Do (sin S) tan (BD-DC tan (AB+AC) tan X tan Σ But (29.), that is, tan} AB-AC) tan į BC 'tan Atap B' tan X tan E X tan A tan X tan 4 (tan 4)". and therefore, tan B as also (tan B)? 'tan Botan (tap B)? tan X tan A (sin D)? But (tan A)? _(sin D) tan A x ; whence tan B tan (sin S) (tan B)2 (sin S) tan B sin D or sin S:sin D :: tan B : tan 4, that is, sin (C+B): sin (C-B) :: tan : BC : tan} (AB - AC); which is the first part of the proposition. tan A cos SX sin D Again, since or inversely tan &_sin S.X cos D tan Σ sin SX cos D tan A cos SX sin Di tan X sin D Xcos D and since tan X tan B sin Sx.cos g; therefore by multiplication, tan B tan $ _(cos D)2 tan A (cos 3)2 tạp X tan $xtan A But it was already shewn that wherefore also tan B (tan B) tan X tan Σ (tan s)? X; tan BẠtan A (tan B)? tan X tan = (cos D2) Now, X °tan B tan Ascos S) as has just been shewn. (cos D2 (tan )? tan Σ Therefore and consequently ,or cos (cos S)2(tan B)?! cos stan B ; and sin S' 3 Х cos D S: cos D :: tan B : tan E, that is cos (C+B): cos (C-B):: tan Cor. 1. By applying this proposition to the triangle supplemental to ABC (11.), and by considering, that the sine of half the sum or half the difference of the supplements of two arches, is the same with the sine of half the sum or half the difference of the arches themselves ; and that the same is true of the cosines, and of the tangents of half the sum or half the difference of the supplements of two arches ; but that the tangent of half the supplement of an arch is the same with the cotangent of half the arch itself; it will follow, that the sine of half the sum of any two sides of a spherical triangle, is to the sine of half their difference as the cotangent of half the angle contained between them, to the tangent of half the difference of the angles opposite to them : and also that the cosine of half the sum of these sides, is to the cosine of half their difference, as the cotangent of half the angle contained between them, to the tangeht of half the sum of the angles opposite to them. Cor. 2. If therefore A, B, C be the three angles of a spherical triangle, a, b, c the sides opposite to them, 1. sin } (A+B): sin } (A - B) :: tanc : tan (à-b). ti a right angled spherical triangle, of the three sides and three anglesa any two being given, besides the right angle, to find the other three. This problem has sixteen cases, the solutions cf which are contained in the following table, where ABC is any spherical triangle Fight angled at A. TABLE for determining the affections of the Sides and Angles found by the preceding rules. 1 AC and B of the same affection, (14). (Cor. 15.) If BC 290° C and B of the same affection, otherwise different, (15.) 2 3 4 AB and C are of the same affection, (14.) If AC and C are of same affection, BC_90°; otherwise BC790°, (Cor. 15.) B and AC are of the same affection, (14.) 5 6 Ambiguous. 7 8 9 When BC_90°, AB and AC of the same ; otherwise of different affection, (15.) | 10 AC and B of the same affection, (14.) 11 When BC 200°, AC and C of the same; otherwise of different affection, (Cor. 15.) | 12 BC_90°, when AB and AC are of the same affection, (1. Cor. 15.) i 13 B and AC of the same affection, (14.) | 14 C and AB of the same affection, (14.) 14 AB and C of the same affection, (14.) | 15 AC and B of the same affection, (14.) 15 When B and C are of the same affection, BC 290°, otherwise, BC 790°, (15.) | 16 The cases marked ambiguous are those in which the thing sougho has two values, and may either be equal to a certain angle, or to thes supplement of that angle. · Of these there are three, in all of which the things given are a side, and the angle opposite to it; and accordingly, it is easy to shew, that two right angled spherical triangles may always be found, that have a side and the angle opposite to it the same in both, but of which the remaining sides, and the remaining angle of the one, are the supplements of the remaining sides and the remaining angle of the other, each of each. Though the affection of the arch or angle found may in all the of cases be determined by the rules in the second of the preceding it is of use to remark, that all these rules except two, may be |