to the included angle. The Author gives in his Work a Table of natural versed sines, which are plainly necessary in his mode of computation. If we use one of the preceding formulæ for computing c, we may, if we please, determine A and B without the aid of Naper's Analogies, and by these expressions, C A, B, two angles, and c the adjacent side are given, and the side a is required. The solution will be deduced from the former, by the aid of the supplemental triangle; A', B', C', a, b, c' being its angles and sides. sin. tan. = sin. (18 1800 3.] B - tan. 2 A+ - 4+B) = sin. (4+B) 2 2 (4 - "') = sin. (3 = 4) = - sin. (4 – 3); * This combination is in the third row (see p. 155,) and therefore not essentially different from the first of that row, namely, (a C b A) which has been already considered. In fact, the two equations involving the four quantities are precisely similar: but, from what was said in p. 157, the same formula of solution cannot suit each case, since in the former, an angle is the quantity required, which in the latter, becomes one of the quantities given. ... generally, sin. (d±5') = ± sin. (1±8). Having now transformed all the terms in [1], [2], (see p. 167), into different expressions, if we substitute the transformed terms in [1], [2], there results These equations [3], [4], expanded into a proportion, are called, [as the former similar ones [1], [2] have been (see p. 169.)] Naper's Analogies. cos. 9.8184449......sin. = 9.8766379 25 33 6......tan. 9.6795032.. = a and b being determined, C may be determined from this Example, or, without the intervening process of finding a and b, by the following method. Determination of the angle C. In the supplemental triangle, by the original form [c], p. 139, cos. c' = sin. d. sin. b. cos. C' + cos. d . cos. b'; or cos. (180oC)=sin. (180o — A). sin. (180o-B). cos. (180°—c) or - and .. 1 cos. C, or, ver. sin. C = 1 + cos. A. cos. B - sin. A . sin. B + ver. sin. c. sin. A . sin. B = 1 + cos. (A + B) + ver. sin. c. sin. A sin. B ; (log. ver. sin.c+log. sin. A+log. sin. B-2 log. cos. (4+B) 2 2. cos.2 (4+ B) sec.? 0; C A+ B = log. cos. 2 +log, sec. 10. C, in the former Example, found independently of a, b. .. log. tan. 0=10.2597736...sec.0=10.3171290 and C = 36° 45′ 27′′ nearly, as before. If we express 1 + cos. (A + B), the versed sine of the supplement of A+ B, by suver. sin. (A + B) we may employ this form for computing C ver. sin. C = suver. sin. (A + B) + ver. sin. c . sin. A. sin. B suver. sin. (A + B). sec.2 0, putting tan.20 sin. A . sin. B. ver. sin.c suver. sin. (A+B) CASE V. (a Ab B). Two sides a, b, and an angle A opposite to one, are given, the angle B, and the remaining angle and side are required. sin. A. sin. b By Cor. 2, page 141, sin. B = sin. a * Since ver. sin. c=2 sin., log. ver. sin. c-log. 2=2 log. sin. C 10 2 A+B 2 ... log. tan. 0=(2 log. sin.+log. sin. 4+log.sin. B-2 log. cos. which form is rather more convenient than the one used in the computation. Z |