opposite parts; and if, in the triangle ABP, co. c is the middle part, BP and AP are the opposite parts. Hence, by (605), {or cos. PC: cos. AP :: sin. (co. a) : sin. (co. c), cos. PC: cos. AP : : cos. a: cos. C. Fourthly. To find the angle A. If, in the triangle BPC, PC is the middle part, co. Cand BP are adjacent parts; and if, in the triangle ABP, AP is the middle part, co. BAP and BP are adjacent parts. Hence, by (604), (617) (618) sin. PC: sin. PA :: cotan. C: cotan. BАР, and BAP is the angle A (fig. 4.), when the perpen- (619) dicular falls within the triangle; or it is the supplement of A (fig. 5.), when the perpendicular falls without the triangle. Fifthly. Bis found by means of (598) sin. c: sin. C::sin.b: sin. B. (620) 37. Scholium. In determining PC, cand BAP, by (615), (617), and (618), the signs of the several (621) terms must be carefully attended to; by means of (496). But to determine which value of B, determined by (620), is the true value, regard must be had to the (622) following rules which will be demonstrated hereafter. I. The greater side of a spherical triangle is always (623) opposite to the greater angle (768). (624) II. Each side is less than the sum of the other two (730). (625) III. The sum of the sides is less than 360° (713). (626) IV. Each angle is less than the difference between 180°, and the sum of the other two angles (822). There are, however, cases in which these conditions are all satisfied by each of the values of (627) B. In any such case this angle can be determined in the same way, in which the angle A was determined by letting fall a perpendicular, from the vertex A on the side BC. But this difficulty can always, (628) by (772), be avoided by letting fall the perpendicular upon that of the two given sides which differs the most from 90°. EXAMPLES. 1. Given, in the spherical triangle ABC, a 45° 54′, b = 138° 32′, and C = 98° 44′; AP = 171°6′ - 138° 32′ 32° 34′. (629) 38. Problem. To solve a spherical triangle, when one of its sides and the two adjacent angles are given. Solution. Let ABC (figs. 4. and 5.) be the triangle ; a the given side, and B and C the given angles. From B let fall on AC the perpendicular BP. First. To find PBC, we know, in the right triangle BPC, the hypothenuse a and the angle C. Hence, by (474), cotan. PBC cos. a tang. C. Secondly. ABP is the difference between ABC and PBC, that is, (630) or (fig. 4.) ABP = В — РВС, (fig. 5.) ABP = PBC — В. 1 Thirdly. To find the angle A. If, in the triangle PBC, co. C is the middle part, PB and co. PBC are the opposite parts; and if, in the triangle ABP, co. BAP is the middle part, PB and co. ABP are the opposite parts. Hence, by (605), (631) or cos. (co. PBC) : cos. (co. ABP) : : sin. (co. C) : sin. (co. BAP), sin. PBC: sin. ABP :: cos. C: cos. BAР ; (632) and BAP is either the angle A or its supplement, as in (619). Fourthly. To find the side c. If, in the triangle PBC, co. PBC is the middle part, PB and co. a are the adjacent parts; and if, in the triangle ABP, co. ABP is the middle part, PB and co. c are the adjacent parts. Hence, by (604), cos. PBC : cos. ABP :: cotan. a: cotan. c. Fifthly. b is found by (598), sin. C: sin. c:: sin. B: sin. b. (633) (634) 39. Scholium. In determining PBC, BAP, and (635) c by (629), (631), and (633), the signs of the several terms must be carefully attended to, by means of (496). To determine which value of b, obtained from (634), is the true value, regard must be had to (623-(636) 626). But if all these conditions are satisfied by both values of b, then b may be calculated by letting fall a perpendicular from C on the side cin the same way in which c has been obtained in the preceding solution. But this case can, by (772), be avoided by letting fall the perpendicular from the vertex of that one of the two given angles which differs the most (637) from 90°. EXAMPLES. 1. Given in the spherical triangle ABC, a = 175° 27', B = 126° 12′, and C = 109° 16'; to solve the triangle. |