= 8032.29 c = 8242.58 A = 71°. 3'. 34" B = 53°. 26'.0",38 log a = 3.9758595, = log c= 3.9160632, L sin A9-9758250, L sin B = 9.9048049, C = 55°. 30.24",16 L sin C = 9.9160287. 107. (3) Having given two sides a, b, and the included angle C, to find the other two angles, and the third side. We have cot which gives (4 – B); and since (A + B) is known, we have ▲ = } (A + B) + } (4 − B) and B = ↓ ( A + B) − ↓ (4 − B) ; and then c is known from the equation 108. Dividing the last equation by equation (1) we find and sin A + sin B = 2 sin (4 + B) cos (4 – B) a formula which will give c by the aid of only two additional logarithms (instead of three which equation (2) requires) since log (a + b) is already employed in the process. 109. When two sides and the included angle are given, the third side c has been found only by means of A and B previously determined. It may however be found directly from the formula which must be adapted to logarithmic computation by means of a subsidiary angle. Among the different transformations which this formula may undergo, the following is the most remarkable. Since C 2 2 = cos C, c2 = (a2 + b′) (cos' € + sino c) – 2 ab (cos C-sin') C = (a + b)' sin' c + (a - b) cos' C The solution only differs from the former one in appearance; for since is equal to (4 – B), this value of c is identical with that furnished by equation (3). 110. If it should happen (as is often the case in practice) that a and b are known only by their logarithms, and that the angles A and B alone are required, then, instead of first finding a and b from the tables which the employment of the formula would require, we may proceed thus. Let & be a subsidiary angle determined by the equation - By this process for finding (4 – B), there are two logarithms fewer employed than if we first determine a and b. 111. (4) Having given the three sides, to find the angles. We have (Art. 93) which gives A, and by similar formulæ may B and C be determined. But we must endeavour to find other formulæ more convenient for logarithms. and substituting for cos ▲ its value, we get successively To simplify this, let half the perimeter of the triangle be noted by s, so that a + b + c = = 28, .. a + b − c = 2 (8 −c), a − b + c = 2 (8 − b), (1). 2 Although the angle be no ambiguity; for A, being the angle of a triangle, is less 113. (2). .(3). As each of these formulæ requires four logarithms for its computation, when we want to find only one angle of a triangle, there is no reason for preferring one to another, except that when is nearly 90°, we must avoid determining it by its sine or tangent, and when it is very small, we must avoid determining it by its cosine; and that when 4 is less than 45°, the tables will determine it from its sine with greater precision than from its cosine; and vice versa when A is greater than 45o (Art. 85—87). When however we want to find two angles, the formula for the tangent of half an angle is to be preferred, as we shall need only the four logarithms of 8, 8-a, s-b, s-c; whereas using either of the others, we shall need six logarithms. 114. A triangle, as we know, cannot be formed with three given sides, unless the sum of any two be greater than the third. This also A _ (s—b)(s−c), appears from the preceding formulæ. Thus, taking sin 2 = bc 2s >2 (a + c), A or s - c> a, and is therefore positive; the value of sin is therefore 2 imaginary; and similarly it may be shewn to be imaginary supposing c> a+b. If a > b + c, then s>b+c; ..s-b>c, s-c>b, .. (s—b) (s - c) > bc, A or the value of sin 2 greater than 1, which is absurd. If b or c be supposed equal to the sum of the two others or to s, the value of A sin is evidently zero; if a = b + c = s, the value of sin 2 and cos 4, what ought to happen, as the triangle is reduced to a straight line. 115. Taking twice the product of the values of sin we find 2 sin A = — ̧ √s (s − a) (5 — b) (s — c) · As this formula requires seven logarithms, it is not convenient for sin A the calculation of A. It shews that is a symmetrical expression a |