(6) Given cos A= and cos B=13, prove that cos C= (7) If sin2 B+ sin2 C=sin2 4, then A=90o. (8) If sin 2B+sin 2C=sin 24, 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+2c3 — a2. (12) Given that a=2 =26, 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+CF2)=3(a2+b2+c2). (16) If D is the middle point of BC, cot ADB= b2-c2 4.S (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). 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. them. Given two sides and the angle opposite one of [Compare Euc. VI. 7.] — 10 = } { log (s — b) + log (s — c) — log s — log (s− a)}. Similarly, 10{log (sc) + log (s - a) - log s-log (s-b)}. 2 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 s, log (s − a), log (8-6), log (8-c). 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=2 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} = =10+{2·2872865+1·9115837 −2·5832555 - 2·0322157} =9.7916995 A whence -31° 45′ 28.5′′; .. A=63° 30′ 57′′. Also B Ltan 10+ (log 81.58+ log 107-70-log 383.05- log 193.77} 2 = -9.5366287=Ltan 18° 59′ 9.8"; .. B=37° 58′ 20′′; C=180o – A – B=78° 30′ 43′′. 255. This Case may also be solved by the formula But this formula is not adapted for logarithmic calculation, and therefore is seldom used in practice. It may sometimes be used with advantage, when the given lengths of a, b, c each contain less than three digits. Example. Find the greatest angle of the triangle whose sides are 13, 14, 15. Let a=15, b=14, c=13. Then the greatest angle is A. (1) Ifa=352.25, b=513.27, c=482.68 yards, find the angle 4, having given log 674.10=2·8287243, log 321·85=2·5076535, log 160.83 2.2063401, log 191-42-2-2819873, (2) Find the two largest angles of the triangle whose sides are 484, 376, 522 chains, having given that log 6.918394780, log 3·15=4983106, log 2.07=3159703, log 1·69=2278867, I tan 36° 46′ 6′′=9.8734581, L tan 31° 23′ 9′′=9.7853745. (3) If a=5238, b=5662, C= =9384 yards, find the angles A and B, having given log 1.0142-0061236, log 4.904='6905505, log 4.486512780, log 7.58='8796692, I tan 14° 38' 9.4168099, L tan 15° 57'-9.4560641, (4) If a=4090, b=3850, C= =3811 yards, find A, having given log 5.87557690448, log 3.85='5854607, log 1.7855-2517599, log 3.811=5810389, L cos 32° 15'9.9272306, L cos 32° 16' 9.9271509. (5) Find the greatest angle in a triangle whose sides are 7 feet, 8 feet, and 9 feet, having given log 3=4771213, L cos 36° 42' = 9.9040529, log 1.4=146128, diff. for 60"= '0000942. (6) Find the smallest angle of the triangle whose sides are 8 feet, 10 feet, and 12 feet, having given that log 2=30103, L sin 20° 42′ = 9.5483585, diff. for 60" 0003342. (7) If a:b:c=4: 5: 6, find C, having given log 2=3010300, log 3=4771213, = L cos 41° 25′ = 9.8750142, diff. for 60" 0001115. == (8) The sides of a triangle are 2, √6, and 1+√3, find the angles. (9) The sides of a triangle are 2, 2 and 3-1, find the angles. |