159. To express the sines, cosines, and tangents of the halfsides of a spherical triangle in terms of the angles of the triangle. Denoting A+B+C by 2 S, so that S is the half-sum of the angles of the triangle, we have B+C−A = 2 (S — A). A). (102) sin b = sin C sin A cos S cos (S- C) (104) sin A sin B Again, adding both members of (C) to unity, we have But A+B − C = 2(S – C), and A − B + C = 2 (S — B). NAPIER'S ANALOGIES. 160. Dividing (99), Art. 158, by (100), we have 161. Multiplying (99) by (100), we have 162. Proceeding as in Art. 157, and applying (111) to the triangle A'B'C', we obtain, But, and sin (A'+ B') tan c' Whence, = † (A' — B') = † (180° — a — 180° + b) = {( − a + b), † (a' — b') = † (180°— A—180°+B)={(−A+B). = tan (90° C) cos (a+b) tan [180° - (A+B)] Therefore, by Arts. 42, 46, and 47, (113) (114) |