8. Show that the ratios of any angle, whether acute or obtuse, are subject to the relations of equality established in Chapter III. 9. How must the relations of inequality of Chapter III. be modified in order to include obtuse angles? § 2. RELATIONS BETWEEN THE SIDES AND THE RATIOS 141. Consider any triangle ABC. The angles at the points A, B, C will be called by the names A, B, C. The sides BC, CA, AB, which are opposite to these angles, will be called a, b, c respectively. 142. To draw the figure by which the ratios of the base angles of a triangle may be indicated. Draw AL perpendicular to BC or BC produced. Then, the angle ACB may be (i) acute, (ii) obtuse, or (iii) right. 143. To prove the double-cosine formula: and ca cos B+ b cos A. 144. To prove the double-sine formula: .. CA sin C = BA sin B, i.e. b sin C = c sin B. COR. Dividing by bc, we have i.e. the sines of the angles are proportional to the opposite 145. By the corollary of the last article, we have fraction the numerator is a length and the denominator a ratio, therefore d will represent a length. 146. To find the sine and cosine of the sum and difference of two angles by means of the double-cosine and double-sine formula. In the equation a = b cos C + c cos B, substitute for a, b, c in terms of d. Thus d sin Ad sin B cos C+ d sin C cos B, .. sin A = sin B cos C + cos B sin C. Similarly sin B [Art. 145. (A.) i.e. (substituting for sin A) sin B sin C cos A+ cos2 C sin B+ cos B cos C sin C, cos A = cos B cos C - sin B sin C. - (B.) = cos (B+ C), [Art. 140. sin (B+ C) = sin B cos C + cos B sin C .....(1), cos (B+C) = cos B cos C-sin B sin C Let B' be the exterior angle at B. Then A = B' - C, sin B = sin B' and cos B sin (B′ – C') = sin B′ cos C – cos B' sin C 147. It should be observed that, by the method of the last article, we have proved the formulæ for the sine and cosine of the sum and difference of two angles, whatever geometrical value the angles considered may have, i.e. whether they are acute or obtuse. Hence, also, the formula which may be algebraically derived from these are universally true for all geometrical angles, viz. the expressions for the tangent and cotangent of the sum or difference of two angles, and the ratios of the double angles, &c. Example. If A, B, C are the angles of a triangle, express sin A+ sin B+sin C as a product of ratios. The given expression =sin {(A+B)+(A − B)} +sin {† (A+B) − † (A − B)} +sin C =2 cos C{cos (A-B)+cos(A+B)}, [sin(A+B)=cos C and sin C=cos (A+B)] =2 cos C. 2 cos A cos B This formula may be written in other forms: thus In fig. (i), (Euc. II. 13), AB = BC2+ CA2-2BC. CL. In fig. (ii), (Euc. II. 12), AB2 – BC2 + CA2 + 2BC . CL. = 150. To prove the sine formula: 2ab sin C = {(a+b+c) (a + b −c) (c + a−b) (b+c-a)}. By the last article, .. 2ab (1 + cos C) = (a2+2ab+b2) - c2 = (a + b + c) (a + b −c) (1), 2ab (1 − cos C) = c2 — (a2 — 2ab+ b2) = (c + a − b) (c− a+b) (2). .. multiplying the above equations (1) and (2), we have 4a2b2 (1-cos2 C)=(a+b+c) (a + b −c) (c + a−b) (b+c-a); .. taking the square root, since 1 – cos2 C = sin2 C, 2ab sin C = √{(a + b + c) (a + b − c) (c + a − b) (b + c − a)}. 151. To prove the half-angle formulæ. Since 1+ cos C = 2 cos2 C and 1 - cos C = 2 sin2 ¿C, substituting in (1) and (2) of last article 4ab cos2 C = (a + b + c) (a + b −c), Dividing, we have (c+a-b) (b+c-a) tan2 C = (a+b+c) (a+b-c) |