62. CASE III.-Having given two sides and the included angle, a, C, b. This case is somewhat similar to the corresponding one in Plane Trig., the base angles being calculated from two separate equations, one giving half the sum and the other half the difference of those angles. The formulæ to be employed have been arrived at on p. 43 (Note to Ex. 8). They have, moreover, been demonstrated geometrically from the relations between the sides and angles of certain right-angled triangles. As they are of fundamental importance in the solution of triangles, we shall now proceed by a direct method to obtain the same results. cos B+ cos C cos A = sin C sin A cos b = x sin C sin a cos b. By addition, (cos A+ cos B)(1 + cos C) = x sin C sin (a + b); (2) by subtraction, (cos B-cos A) (1 - cos C) = x sin C sin (a - b). (3) On taking the positive sign in (1), and dividing by (2), On taking the negative sign in (1), we obtain, by a similar Dividing each of these equations by results in equation (1) to eliminatex, we have formulæ supplemental to (a) and (B'), viz., Substituting this value in (2), we have cos (4 – B) sin = sin § (a + b) sin § C. Similarly, (B) с sin (A + B) cos = cos(ab) cos C, (2) and 2 sin (A – B) sin = sin § (« – b) cos § C. (8) 65. It will be thus seen that when each of Napier's Analogies is added to or subtracted from unity, the results are easily transformed into Delambre's formulæ. It is also to be noticed that if the equation cos (A + B) cos c = cos(a + b) sin be applied to the co-lunar triangle of parts a, sin C (AB) sin c = sin(b) cos C. And by a similar method, the equation gives sin (A+B) cose cos (a - b) cos C cos (A – B) sin c = sin (a + b) sin C. b, π-c, Remark. From the three preceding Articles it has been shown that Napier's four Analogies are reducible, by aid of the polar triangle, to two independent theorems, involving five parts of a spherical triangle; and that Gauss's four formulæ are, by aid of the colunar triangle, likewise reducible to two independent theorems. I 66. In the solution of triangles, CASE III., the for mulæ (A + B) and (A – B), from which enable us to calculate we obtain A and B. The remaining element c is determined from the equation cos (A + B) cos c = cos(a + b) sin C ; or, as will be hereafter shown, it may be determined from the formula cos c = cos a cos b + sin a sin b cos C, without having previously determined 1 or B. (A + B) = 56° 29′ 17′′, and (4 – B) = 27° 41′ 03′′. |