Two sides and the included angle given. 51. By using table XXIII, the following rule is obtained for finding the third side, when two sides and the included angle are given. Add together the log. Rising of the given angle, and the log. sines of the two given sides. The sum is the logarithm of a number, which is to be subtracted from the natural cosine of the difference of two given sides (regard being had to the sign of this cosine). The difference is the natural cosine of the required side. 52. EXAMPLES. 1. Calculate the value of log. Ris. of 4h 28m. 2. Given in the spherical triangle two sides equal to 45° 54' and 138° 32', and the included angle 98° 44'; to solve the triangle. Two sides and the included angle given. II. The third side is thus calculated by means of (308), −0.99701-Nat. cos. (138° 32'+45° 54′)= N. cos. 184° 26' -0.59369—Nat. cos. 126° 25′ 10′′ — c. III. The third side is thus calculated by § 50. 3. Calculate the log. Ris. of 11h 12m 20o. Ans. 5.29632. . 4. Given in a spherical triangle two sides equal to 100°, and 125°, and the included angle equal to 45°; to solve the triangle. Ans. The third side=47° 55′ 52′′ The other two angles 69° 43′ 48′′, and 128° 42′ 48′′. = A side and the two adjacent angles given. 53. Problem. To solve a spherical triangle, when one of its sides and the two adjacent angles are given. [B. p. 438.] Solution. Let ABC (figs. 32 and 33) 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 Napier's Rules, cotan. PBC=cos, a tang. C. (312) Secondly. ABP is the difference between ABC and PBC, that is, (fig. 32) ABP = B− PBC, or (fig. 33) ABP = PBC — B. (313) 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 Bowditch's Rules, cos. (co. PBC): cos. (co. ABP)=sin. (co. C): sin.(co. BAP), or sin. PBC: sin. ABP = cos. C: cos. BAP; (314) and BAP is either the angle A or its supplement. 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 Bowditch's Rules, cos. PBC: cos. ABP = cotan. a: cotan. c. (315) A side and the two adjacent angles given. Fifthly. b is found by the proportion sin. C sin. c — sin. B : sin. b. (316) 54. Scholium. In determining PBC, BAP, and c by (312), (314), and (315), the signs of the several terms must be carefully attended to, by means of Pl. Trig. $61. To determine which value of b, obtained from (316), is the true value, regard must be had to the rules of § 46. But if all these conditions are satisfied by both values of b, then b may be calculated by letting fall a perpendicular from Con the side c in the same way in which c has been obtained in the preceding solution. But this case can be avoided by letting fall the perpendicular from the vertex of that one of the two given angles, which differs the most from 90°. C 55. Corollary. Since 180°. - a, 180° — b, and 180° are the angles of the polar triangle, and 180°—A, 180°—B, and 180°C are its sides; we have given in the polar triangle the two sides 180°- B, and 180° - C, and the included 180°-a; so that the polar triangle might be solved by $45. 56. Corollary. If formula (307) is applied to the polar triangle of the preceding section, it becomes by Pl. Trig. $61, or cos. A = cos. B cos. C—sin. B sin. C cos. ɑ, cos. A =- cos. B cos. C+ sin. B sin. C cos, a. (317) 57. Corollary. In the same way (308) becomes by (99) and (123), cos. A —— cos. (B+C) — 2 sin. B sin. C (sin. a)2, (318) |