dicular to OB and OC respectively. Join L and M; join L and N. Then LML OB, and LN 1 OC. (Geom.) 182. If A'B'C' be the polar triangle of ABC, then (Fig. 67) Relation involving the angles and one side. Solving these equations for cos a, cos b, cos c, we have cos Acos B cos C (4) cos b = ; (5) sin C sin A cos a = cos b cos c + sin b sin c cos A. Substituting in this formula the value of cos c obtained from (1), we obtain or cos a (1 — cos2 b) = sin a sin b cos b cos C + sin b sin c cos A; and similarly, cos a sin b = = sin a cos b cos C+ sin c cos 4; cos b sina sin b cos a cos C+ sin c cos B; (6) (6) is a relation involving two angles and the sides. By treating (3) similarly, (7) is a relation between the angles and two sides. (8) cot a sin b = cos b cos C+ sin C cot A. The student may derive the five corresponding formulæ : 184. To express cos▲, sin A, tan ▲ in terms of the sides. If we put a + b + c = 2 s, (a) may be written: cos C V sins sin(s- b) sin a sin c sin s. sin(s (Art. 92.) (Art. 75.) (a)(Art. 82.) (9) Using the results of (5), 185. To find cosa, sina, tana. the student may show that, if S = } (A + B + C), 2 = sin B sin C 2 sina cosa (13) (14) (Art. 91.) cos S cos (SA) cos (S — B) cos (S – C). (15) 186. I. Since any side a < 180°, and any angle A< 180°, cosa, sina, tana, sin A, cos 1⁄2 A, tan ↓ A are all positive; hence the sign of the radical in expressions (9)–(15) is positive. II. Since 8< 180°, a < 180°, b<180°, c<180°, and since the differences s — a, s — b, s — c are less than 180° and positive, the expressions (9) - (11) are real. III. If a', b', c' are the sides of the polar triangle of A, B, C, then a' (b'+c'); Since 270° > S°> 180°, cos S is negative; i.e. - cos S is positive. Therefore the expressions (12) — (15) are real. 187. cos (A + B) = cos | A cos B – sin A sin † B. (Art. 75.) Substitute in this equation the values of cos A, etc., from (9) and (10). sin s sin (sa) sin (s cos (A+B)= sin c sin b sin a sin (sc), sin c |