2. If A=55°, B=65°, c=270, find a; given log 2, log 3, 3. If A=45° 41′, C=62° 5′, b=100, find c; given 4. If B=70° 30′, C=78° 10′, a=102, find b and c; given log 2=301, log 102-009, log 1.85=267, log 1.92=283, 5. 6. 7. L sin 70° 30′ =9.974, Lsin 78° 10′=9.990, L sin 31° 20'=9.716. If a=123, B=29° 17′, C=135°, find c; given log 2, If A=44°, C=70°, b=1006·62, find a and c; given = If a=1652, B=26° 30′, C=47° 15′, find b and c; L sin 73° 45'-9.9822938, log 1.652 2180100, L sin 26° 30' 9'6495274, log 7.6780-8852481, D=57, L sin 47° 15′-9.8658868, log 1.2636=1016096, D=344. 192. To solve a triangle when two sides and the angle opposite to one of them are given. Let a, b, A be given. Then from sin B: b =- sin A, we have log sin B-log b―log a+ log sin A; whence B may be found; then C is found from the equation C=180° – A – B. .. log c=loga+log sin C-log sin A. If a<b, and A is acute the solution is ambiguous and there will be two values of B supplementary to each other, and also two values of C and c. [Art. 147.] Example. If b=63, c=36, C=29° 23′ 15′′, find B; given log 2=3010300, log 7=8450980. L sin 29° 23′-9.6907721, diff. for 1'=2243, L sin 59° 10′9.9338222, diff. for 1'=755. EXAMPLES. XVI. e. 1. If a=145, b=178, B=41° 10′, find A; given log 178=2.2504200, L sin 41° 10' 9.8183919, log 145=2.1613680, L sin 32° 25′ 35′′=9′7293399. 2. If A = 26° 26′, b=127, a=85, find B; given log 1.27=1038037, L sin 26° 26'=9.6485124, log 8.5 9294189, L sin 41° 41′ 28′′-9.8228972. = 3. If c=5, b=4, C=45°, find A and B; given log 2, L sin 40°=9·8080675, L sin 51° 18′=9·8923342, 5. If B=112° 4′, b=573, c=394, find A and C; given L sin 39° 35′ = 9.8042757, diff. for 60′′=1527, 6. If b=84, c=12, B=37° 36′, find A; given log 7=8450980, L sin 37° 36′ = 9.7854332, 7. Supposing the data for the solution of a triangle to be as in the three following cases, point out whether the solution will be ambiguous or not, and find the third side in the obtuse angled triangle in the ambiguous case: log 6.0389=7809578, L sin 38° 41'=9.7958800, 193. Some formula which are not primarily suitable for working with logarithms may be adapted to such work by various artifices. 194. To adapt the formula c2=a2+b2 to logarithmic compu tation. Since an angle can always be found whose tangent is equal to a given numerical quantity, we may put obtain =tan 6, and thus α c2=a2 (1+tan2 0) = a2 sec2 0 ; The angle is called a subsidiary angle and is found from the equation log tan = log b - loga. Thus any expression which can be put into the form of the sum of two squares can be readily adapted to logarithmic work. H. K. E. T. 12 195. To adapt the formula c2=a2+b2-2ab cos C to logarithmic computation. C C =(a2+b2-2ab) cos2 +(a2+b2+2ab) sin2 C where is determined from the equation log tan 8=log (a+b) — log (a - b) +log tan. 196. When two sides and the included angle are given, we may solve the triangle by finding the value of the third side first instead of determining the angles first as in Art. 189. Example. If a=3, c=1, B=53° 7′ 48′′ find b; given log 23010300, log 2.5298=4030862, diff. for 1=172, L cos 26° 33′ 54′′-9.9515452, L tan 26° 33′ 54′′9.6989700. |