But if AD be without the triangle, BAD-CAD-BAC, and therefore sin. (AB+AC): sin. (AB-AC) :: cot. (BAD+CAD): tan. BAC; or because cot. (BAD+CAD): tan. BAC :: cot. & BAC : tan. (BAD+CAD), sin. (AB+AC): sin. (AB-AC) cot. BAC : tan. (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. Let A and B be two arches, tan. A+tan. B : tan. A~tan. B :: sin. (A+B): sin. (A-B). For, by 6.page 243, sin. Axcos. B+cos. Axsin. B=sin. (A+B), and therefore dividing all by cos, A cos. B, H sin. (A+B), that is, because sin. A cos. Axcos. B cos. A sin. A + sin. B cos. A cos. B =tan. A, tan. A+tan. B In the same manner it is proved that tan. A―tan. B Therefore tan. A+tan. B: tan. A-tan. B :: sin. (A+B): sin. (A-B). Q. E. D. PROP. XXXI. The sine of half the sum of any two angles of a spheri cal 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. (BDCD), sin. 28: sin.. 2D tan. B: tan. X. Now, sin. 28=sin. (S+S)=2 sin. S×cos. S, (Sect. III. cor. Pl. Tr.). In the same manner, sin. 2D=2 sin. Dx cos. 1). Therefore sin. Sxcos. S: sin.D× cos. D☀ tan. B : tan. X. A = Again, in the spherical triangle ABC it has been proved, that sin. C+siu. B; sin. C-sin. B: sin. AB+sin. AC sin. AB-sin. AC, and since sin. C+sin. B-2 sin. (C+B)+cos. & (C-B), (Seet. III. 7. Pl. Tr.) 2 sin. Sxcos. D; and sin. C-sin. B=2 cos. (C+B) ×sin. (C-B)=2 cos. Sxsin. D. Therefore 2 sin. Sxcos. D: 2 cos. Sx 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 (ABAĆ) :: tan : tan ▲, Σ being equal to † (AB+AC) and ▲ to (AB-AC). Therefore sin Sxcos D: cos S×sin D :: tan Σ : tan A. Since then tan X sin DXeos D cos Sxsin D ; and tan B sin Sxcos S sin Sxcos D (sin D)2 ×cos Sxcos D sin S) Xcos SXcos D tan (AB+AC), that is, tan X tan A tan A tan Σ ,bymultiplying equals (sin D) (sin S) tan X tan Σ 'tan A tan B (tan A) by equals, But (29.) and therefore, tan B tan Σ tan BC But tan Xtan A_(sin`D)”. X tan B tau Σ (sin S) sin D sin S' as also tan X tan A whence (tan ▲) __ (sin D) or sin S: sin D: tan B: tan▲, that is, sin (C+B): sin (C-B): tan BC tàn (AB-AC); which is the first part of the proposition. tan X sin Dxcos D, therefore by multiplication, But it was already shown thattan tan Extan, wherefore also and consequently cos D eos S tan B' S: cos. D:: tan. B: tan. E, that is cos. (C+B): cos. (C−B) : ; tan. BC: tan. (C+B); which is the second part of the proposition. Therefore, &c. Q. E. D. COR. 1. By applying this proposition to the triangle supplemental to ABC (11. 5.), 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 tangent 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. I. sin. II. cos. III. sin. 1V. cos. (A+B): sin. (A-B) :: tan. c: tan. (a-b). PROBLEM I. In a right angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle, to find the other three. This problem has sixteen cases, the solutions of which are contained in the following table, where ABC is any spherical triangle right angled at A. SOLUTION. 23 1 2 3 4 5 7 8 9 R : sin. BC : : sin. B: sin. AC, (19). 23 15 23 15 (20 16 LI B TABLE for determining the affections of the Sides and Angles found by the preceding rules. AC and B of the same affection, (14). If BC 90°, AB and B of the same affection, otherwise dif ferent, If BC90° C and B of the same affection, otherwise dif ferent, AB and C are of the same affection, 1 (Cor. 15.) 2 (15.) 3 (14.) 4 (Cor. 15.) 5 6 689 If AC and C are of the same affection, BC90°; otherwise B and AC are of the same affection, Ambiguous. Ambiguous. Ambiguous. When BC90°, AB and AC of the same; otherwise of AC and B of the same affection, BC90°, when AB and AC are of the same affection, (15.) 10 (14.) 11 When B and C are of the same affection, BC90°, otherwise, BC 790°, The cases marked ambiguous are those in which the thing sought has two values, and may either be equal to a certain angle, or to the 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 show, 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 other cases be determined by the rules in the second of the preceding tabies, it is of use to remark, that all these rules except two, may be reduc |