249. The student is advised to make himself thoroughly familiar with the following formulæ : (3) a cos A +b cos B-c cos C=2c cos A. cos B. (7) a sin (B-C)+b sin (C − A)+c sin (A − B)=0. (11) a+b+c=(b+c) cos A+ (c+a) cos B+(a+b) cos C. (12) b+c-a=(b+c) cos A − (c− a) cos B+(a−b) cos C. (14) a (b2+c2) cos 4+b (c2+a2) cos B+c (a2+62) cos C-3abc. (15) a cos (A+B+C)-b cos (B+A)-c cos (A+C)=0. ** MISCELLANEOUS EXAMPLES. LXV. (1) If p is the length of the perpendicular from A on BC, (6) Given cos A= and cos B=13, prove that cos C = − }. (7) If sin2 B+ sin2 C=sin2 A, then A=90o. (8) If sin 2B+sin 2C=sin 2A, then either B=90° or C=90o. (9) If A: B: C=1: 2: 5, then 1+4 cos A. cos B. cos C=0, and a2, 62, c2 are in A.P. (11) If D is the middle point of BC, prove that 4AD2=262+2c2 - a2. (12) Given that a=2 -2b, and that A=3B, prove that C=60o. (13) abc (a cos A+ b cos B+c cos C)=8,S2. (15) If D, E, F are the middle points of the sides BC, CA, AB, then 4 (AD2+BE2+CF12) = 3 (a2+b2+c2). (16) If D is the middle point of BC, cot ADB= = b2-c2 (17) If d, e, f are the perpendiculars from A, B, C on the opposite sides of the triangle, then a sin A +b sin B+csin C=2 (d cos A+e cos B+fcos C). [Some of the Examples in the Appendix might be worked by the student at this stage.] CHAPTER XVII. ON THE SOLUTION OF TRIANGLES. 250. The problem known as the Solution of Triangles may be stated thus: When a sufficient number of the parts of a triangle are given, to find the magnitude of each of the other parts. 251. When three parts of a Triangle (one of which must be a side) are given, the other parts can in general be determined. There are four cases. I. Given three sides. II. Given one side and two angles. [Compare Euc. I. 8.] [Euc. I. 26.] [Euc. I. 4.] III. Given two sides and the angle between them. IV. Given two sides and the angle opposite one of 253. In practical work we or, A L tan 2 (8 − b ) ( s − c) 8 (8 - a) 4 - 10 = } { log (8 − b ) + log (s — c) – log s — log (s — a)}. Similarly, (8b) -- 8 10={log (sc) + log (8 − a) - log s - log (8 - b)}. convenient as the tan formula, when one of the angles only is to be found. If all the angles are to be found the tangent formula is convenient, because we can find the L tangents of two half angles from the same four logs, viz. log 8, log (s — a), log (s—b), log (sc). To find the L sines of two half angles we require the six logarithms, viz. log (sa), log (s—b), log (sc), log a, log b, log c. Example. find A and B. Given a=275.35, b=189.28, c=30147 chains, Here, s=383 05, s- a=107·70,s-b=193·77, s-c=81.58. = = 10 + † {log 193·77+log 81.58 - log 383 05 - log 107.70} |