sin B = cos AC cos C; taking the complements of B and C, cos B = cos AC sin C; Here the circular parts all lie together, and C being in the middle, is the middle part, and BC and AC the extremes disjunct. sin C tan BC tan A C; taking the complements of the hypothenuse and of angle C, QUADRANTAL TRIANGLES. 39. Quadrantal triangles can be solved by the same rules as right-angled triangles for using the polar triangle; we see that since one side is a quadrant, and that in the polar triangle A' 180 = a; .. A' = 180° 90° 90. In the polar triangle, since A'90, we have by the equa tions, page 10, A' = 180° — a; B′ = 180° — b; c' = we get these results, 180° — C; can deduce all the cases of quadrantal triangles. Given A B=c= 32° 57′ 6′′ and A C=b= 66° 32′, to 9.8258956, which is the cosine of 47° 57′ 16′′, but since cos A is negative, A must be greater than 90°. From half the sum of the three sides subtract each of the two sides which contain the required angle. Add the log sines of these two remainders, and the complement logs of the sines of the sides which contain the angle. Half the sum of these four logarithms will give the log sine of half the required angle. Thus: 79° 17′ 14′′ 110 58 2)247 17 14 123 38 37 = sum of the three sides. By a similar operation the angles B and C may be found; but when one angle is known, the other two are easily determined by Art. 13, page 6. CASE 2. Given the angle A = 32° 20′ 30′′, the side b 72° 10′ 20′′, and the side a 78° 59′ 10′′, to find B, C and c. = log sin Blog sin A+ log sin blog sin a log cot C = log tan (A+B) + log cos (a + b) 9.3962727 log cos(a - b) = 9.7217470 but we can find it directly from Napier's Analogies. |