The equation c2=a+b2 directly determines c, but as the operation is somewhat laborious it may be facilitated in the following manner. = Since c2=a2+b2=a2 (1+ hence, c= b2 tan 0, then c2-a2 (1+tan 20)=a2 sec 20 α : cos now, log tan log b+10-log a, from which, by computation, we find 0=2° 30′ 36′′, and thence, cos ◊ =9.9995833. Then log clog a+10-cos : from this (48.) The sides of a plane triangle are proportional to the sines of the opposite angles. Let A B C be a plane triangle (fig. 8.), and B D the perpendicular from B upon the side A C; then by (47.), B D=A B . sin A, also B D=B C . sin C ; for if the angle A C B is obtuse, sin B C D sin ACB (13). Hence, generally, in the triangle A B C, A B . sin A=B C . sin C, or AB: BC:: sin C: sin A. 1 (49.) To find the cosine of an angle when the three sides are given. When the triangle A CB is acute angled, AB-A C2+B C2-2 AC. CD, (Euc. 11. 13); and, when it is obtuse, C being the obtuse angle, AB-A C2+BC2+2 AC. CD. (11. 12.) In the first case, C D=B C. cos C'; in the second, C D=B C . cos BCD hence, generally, =-BC.cos ACB (13.); A B2 A C2+BC2-2 AC. BC. cos C. This form is not adapted to logarithmic computation. (50.) Resuming the equation, cos Ca2+b2—c2 we have, 1+cos C, or (23.), 2 ab Let a+b+c=2 S, then a+b-c-2 (S—c), hence, 2 cos2 C_28.2 (S-c) 2 S 2 α b , or, (51.) Also, 1—cos C, or 2 sin2=1 2 a2+b2-c2 2 ab _c2 — (a—b)2 _ (c+a−b). (c+b—a) (53.) Multiply the product of the same equations C • cos2, or (23.), 4 S. (S-a). (S—b) · (S—c) D All these forms are convenient for calculation by logarithms. (54.) Given the two sides (a, b), and the included angle (C), to find the other angles and the third side. From (47), a b :: sin A : sin B; hence, a+ba-b sin A+ sin B: sin A-sin B, cot ja 2 a+b 2 4-B being known, their sum gives the 2 value of A, and their difference that of B. The third side (c), may be determined by (48). (55.) It is, however, frequently desirable to obtain the third side without the previous process of finding the angles; and this may be accomplished by the following method. By (49.), c2 a2+b2-2 a b cos C a2+b2+2 a b-2 ab-2 a b cos C = (a+b)2 — 2 a b (1+cos C) - C the second and add the results, we shall then have, c2 (sin2+cos2), or (16.), C c2=(a−b)2 cos2 +(a+b)2 sin2 C 2 2 2 |