(ii.) cos A, or (iii.) tan A, then there is only one value of A, which value can be found from the Tables. the value of any one of its ratios is given. Similar remarks of course apply to the angles B and C. and Now EXAMPLE 1. To prove sin (A + B) = sin C. A+B+C 180°; .. A + B = 180° — C, = and 2 EXAMPLES. XXVI. Find A from each of the six following equations, A being an angle of a Prove the following statements, A, B, C being the angles of a triangle: From A, any one of the angular points, draw AD perpendicular to BC, or to BC produced if necessary. There will be three cases. Fig. i. when both B and C are acute angles; Fig. ii. when one of them (B) is obtuse; Fig. iii. when one of them (B) is a right angle. Then, = b cos C+ c cos B. [For, cos B : Similarly it may be proved that, = cos 90° = 0.] bc cos A+ a cos C; c = a cos B+ b cos A. 106. III. To prove that, in any triangle, the sides are proportional to the sines of the angles opposite; or, To prove that From A, any one of the angular points, draw AD perpendicular to BC, or to BC produced if necessary. Then, I. Fig. 50. AD=b sin C; for, = sin C [Def.]; AD Take one of the angles A. acute. sin A sin B a sin A = b sin B = с sin C a2 = b2 + c2 · 2 bc cos A. Q.E.D. Then of the other two, one must be Let B be an acute angle. From C draw CF perpendicular to BA, or to BA produced if necessary. There will be three figures according as 4 is less, greater than, or equal to a right angle. Then, or, or, CA2+ AB2-2. BA FA; a2 = b2 + c2 2c. FA (Geom.) = b2+c2-2 cb cos A. (For FA= bcos A.) II. Fig. 51. BC2 = CA2 + AB +2. BA AF; a2 = b2 + c2 + 2 cb cos FAC (Geom.) ** 109. The formulæ of Art. 108 may be obtained directly from those of Art. 105. Multiplying (1), (2), (3) by a, b, c respectively and adding, we obtain a2 + b2 + c2 = 2 a (b cos C+c cos B) + 2 bc cos A = 2 a2 + 2 be cos A. EXERCISE I. Find the two corresponding expressions, viz., for cos B and cos C. EXERCISE II. If a = 5, b = 6, c = 7, find cos A. and 110. VI. Let s stand for half the sum of a, b, c; so that (a+b+c)=2s. Then, (b+c − a) = (b+c+a-2a)=(28-2a)=2(sa), (c+ab)=(c+a+b2b) = (2 s−2b)=2(s — b), |