These values being of great importance, we shall collect them for easy reference. A 2 sin 4 = 2 √/{S (S − a) ( S − b) (S −c)} ... ... ... ... ... . (a), bc sin § 4 = √ √ (S – b) (S – c) cos § 4 = tan bc ..... By the same method as above, sin B and sin C may be determined; but they may be more easily found from equation (4) thus, put B for A; a for b and b for a, and we have 2 sin B ̧ √{S (S− a) ( S − b) (S − c)} . . . . . . (e). ac Also putting c for A, a for c, and c for a, we have sin C = 2 √/{S (S − a) ( S − b) (S — c)} · · · · · · (ƒ). PRACTICAL OBSERVATIONS. Equation (a) may be used in all cases where A does not approach near to 90o. Equation (b) may be used in all cases where A approaches near to 90o. Equation (c) may also be used in all cases where A approaches near to 90°. Equation (d) may be used in all cases where A does not approach near to 180°. also Where A represents the area of the triangle. We may express the area of a triangle in terms of the sides from the above. 2 A == 24 AB.AC. {S (S − a ) ( S − b ) ( S − c)} bc · b ̧ √ {S ( S − a ) ( S − b) (S – c)} 19. When two sides and their included angle are given, But by Art. 17, sin A+ sin B=2 sin & (A + B) cos & (A – B), and sin Asin B = 2 sin (AB) cos (A + B); log tan (A – B) = log (a - b) — log (a + b) + log cot C. log clog a + log sin C-log sin A ; c may also be found from the formula, 4 ab now since (-b), sin* 4 C = {(2a1 60 in 4 CJ, (a α which, being a square, is necessarily a positive quantity, and as this may be of any magnitude whatever, we may put then c(a - b)(1 + tan3 0) = (a - b)2 sec2 0; which is adapted for logarithmic computation, thus: log c=log (a - b) + log sec - 10. The following is useful in cases where a and b are nearly equal: c2 a2 + b2 − 2 ab cos C ; = but cos C = 2 cos2 & C - 1 ; '. c2 = a2 + b2 — 2 ab (2 cos3 & C - 1) = a2 + b2+2ab-4ab cos2 & C = (a + b)2 — 4ab coso & C' = (a + b)a { 1 Now 4 ab cos2 C (a + b)2 }. 4ab cos C is always less than unity, (a + b)3 for Jab is less than (a+b)*, or (a+b) is greater than 4 ab, and since the cosine cannot exceed unity, it is evident that the above is a proper fraction, and we may put 4 ab cos2 C (a + b)2 ; .. c2 = (a + b)2 (1 - sin3 0) = (a + b)2 cos2 0, 1 2 2 1 log sin = log 2+ log a log b + log cos C - log (a+b). ON THE AMBIGUOUS CASE. 20. When two sides of a triangle and the angle opposite one of them are given, there is evidently an ambiguity; for if CA be taken less than BC, but greater than a perpendicular from C on BA or BA produced, and if with C as centre and CA as radius, a circle be a+b ·>√√ab or a+b>2√√/ab or a2+2ab+b2>4ab, for any even power of a quantity is positive. described, it must necessarily cut the line BAA' in two points, while in all other cases it will either fall short of it or touch it only in one point. When AC is less than a sin B, the triangle is impossible: but when AC = a sin B, the triangle is right angled, and there is no ambiguity. And when AC is greater than BC, there can be only one triangle, and, if it fall between these limits, there is an ambiguity, for there will be two triangles having the same data; thus when the least side AC is opposite the given angle B, it is evident that either ABC or A'BC may be the triangle required. 21. To find the area of a triangle in the terms of the radius of the inscribed circle, and sides of the triangle. The triangle ACB is composed of the triangles AOB, AOC, and BOC; O being the centre of the inscribed circle. Now the area of the triangle AB. OD cr C AOB = a+b+c Since = 2 S, the semi-perimeter, we may express the radius in terms of the sides; putting, as before, ▲ for the area of the triangle, A A=rS; ..r= ..(1) ; but A={S(S-a) (S-b) (S−c)} by p. 56; |