Then, by substitution and extraction of the square root, cos B =√sin s sin (s—b) csc a csc c tan B = √cscs csc (s—b) sin (s — a) sin (s—c) If we place (A+B+C)=S, and proceed in the same manner as before, we obtain the following results : S tana = √—cos $ cos (S — A) sec (S — B) sec (S — C) [48] tanb √cos S cos (S—B) sec (S—A) sec (S-C) sinc= √cos S cos (S-C) csc A csc B cos (S — A) cos (S — B) csc A csc B cos c=√cos tanc= cos S cos (S — C) sec (S — A) sec (S — B § 55. Gauss's EQUATIONS AND NAPIER'S ANALOGIES. By § 27 [5], cos (A+B)=cos A cos B-sin A sin B; 1 1 or, by substituting for cos A, cos B, sin A, sin B, their values given in § 54, and reducing, This value, by applying §§ 29 [12], 31 [21], and observing that the expression under the radical is equal to sin C, becomes 2 sin c cos (8) sin C; cos(A+B)= 2 sinc cos c and this, by cancelling common factors, clearing of fractions, and observing that sc=(a+b), reduces to the form cos (A+B) cos c = cos(a+b) sin & C. By proceeding in like manner with the values of sin (A+B), cos1⁄2 (A−B), and sin1⁄2 (A — B), three analogous equations are obtained. The four equations, cos (A+B) cos c = cos(a+b) sin 1 C are called Gauss's Equations. [49] By dividing the second of Gauss's Equations by the first, the fourth by the third, the third by the first, and the fourth by the second, we obtain There will be other forms in each case, according as other elements of the triangle are used. These equations are called Napier's Analogies. In the first equation the factors cos(ab) and cot Care always positive: therefore, tan (A+B) and cos (a+b) must always have like signs. Hence, if a+b180°, and therefore cos(a + b) > 0, then, also, tan (A + B) > 0, and therefore A + B < 180°. Similarly, it follows that if a+b> 180°, then, also, A+B> 180°. If a+b=180°, and therefore cos (a+b) = 0, then tan (A + B) = ∞ ; whence (A+B)=90°, and A+B=180°. Conversely, it may be shown from the third equation, that a+b is less than, greater than, or equal to 180°, according as A+B is less than, greater than, or equal to 180°. § 56. CASE I. Given two sides, a and b, and the included angle C. The angles A and B may be found by the first two of Napier's Analogies; viz.: cos(a-b) tan (A+B) cot C. After A and B have been found, the side c may be found by [44] or by [50]; but it is better to use for this purpose Gauss's Equations, because they involve functions of the same angles that occur in working Napier's Analogies. Any one of the equations may be used; for example, from the first we have If the side c only is desired, it may be found from [45], without previously computing A and B. But the Formulas [45] are not adapted to logarithmic work. Instead of changing them to forms suitable for logarithms, we may use the following method, which leads to the same results, and has the advantage that, in applying it, nothing has to be remembered except Napier's Rules: B Make the triangle (Fig. 45), as in § 53, equal to the sum (or the difference) of two right triangles. For this purpose, through B (or A, but not C) draw an arc of a great circle perpendicular to AC, cutting AC at D. Let BD—p, CD—m, AD=n; and mark with crosses the given parts. By Rule I., cos C tan m cot a, whence tan m = tan a cos C. a A By Rule II., cos a = cos m cos p, cos c = cos n cos p, Therefore, cos c sec n = cos a sec m ; whence cos pcos a sec m. It is evident that c may be computed, with the aid of logarithms, from the two equations tan m tan a cos C, COS C = = cos a sec m cos (b—m). EXAMPLE. Given a 97° 30' 20", b=55° 12'10", C=39° 58'; 1. Write formulas for finding, by Napier's Rules, the side a, when b, c, and A are given, and for finding the side b when a, c, and B are given. |