By (C). log. sin. == log. R+; {log. (s—a) + log. (s—b)}—¿{log.a+log.b} log. (s—a) = 1-4996871}{log.(s—a)+log.(s—b)} = 1·4210549 log. (s—b) 1·3424227 S log. R = 10. 44° 41′ 10′′ C = 89° 22′ 20′′ = log. R+&{log.(s—a)+log.(s—b)}— { {log.s+log.(s—c)} 1·4996871 log. (s—b) = 1.3424227 2-8421098 log. s = 1.8102325 log. sc = 1.0413927 § .. {log. s + log. (s—c)} = 1.4258126 Subsidiary Angles are angles which, although not immediately connected with a given problem, are introduced by the computist in order to simplify his calculations. Their use, and the method in which they are employed, will be understood from what follows. When two sides of a triangle, and the included angle, are given, according to the method pursued in the last chapter, we must determine the two remaining angles before we can compute the third side. It frequently happens, however, in practice, that the side only is required, and it therefore becomes desirable to have some direct method of computing the side independently of the two angles. Suppose that a, b, C are given, and c is required. By chap. III. prop. 4, c2 = a2 + b2 — 2 ab cos. C the side c is determined theoretically at once by this expression, but the formule is not adapted to logarithmic computation, and would, if employed practically, lead to a very tedious and complicated calculation. We can, however, put this expression under a form adapted to logarithmic calculation, by having recourse to an algebraical artifice, and introducing a subsidiary angle. = a2 + b2 — 2 ab cos. C Adding and subtracting 2 ab on the right hand side. c2 = a2 + b2 - 2 ab + 2 ab - 2 ab cos. C log. tan. = log. 2 + § (log. a + log. b) + log. sin. C - log. (a - b) being thus determined, log. sec. 7 can be found from the tables, and the value of c becomes known. The angle o̟, which is introduced into the above calculation, in order to render the expression convenient for logarithmic computation, is called a subsidiary angle. The above transformation may be effected in a manner somewhat different, as before. Assume c = a + b 2 ab cos. C = a2 + b2 + 2 ab 2 ab 2 ab cos. C log, clog. (a + b) + log. cos. 4 — log. R Q Q As before the angle & must be determined from the equation. that 2ab is always less than (a + b), this is easily done. But since (a—b) is necessarily a positive quantity, it must always be greater than 0 (except in the particular case ab, where it is = 0), and therefore C 2 ab cos. 2 is always less than unity, and consequently an angle may (a + b) always be found whose sine is equal to it. In solving the same case of oblique-angled triangles, we determined the difference of the angles A, B from the equation. In the solution of certain astronomical problems, the logarithms of the sides a, b are given, but not the sides themselves, and these logarithms being given, The angle A → B 2 thus becomes known from the logs. of a and b, without calculating a and b. In the same way we may have, C A B 2 A GREAT variety of geometrical problems may be solved with much elegance by the introduction of trigonometrical formulæ. We shall give a few examples PROB. I. To express the area of a plane triangle in terms of the sides of the triangle. Let CD be a perpendicular from C upon AB. bc 2 bc sin. A 2 2.bc B =2·5c⋅ √s (8 — a) (s — b) (s — c) ... Chap. 1'I =Vs (sa) (s—b) (s—c) PROB. II. To express the radius of a circle inscribed in a given triangle, in terms of the sides of the triangle. To express the radius of a circle circumscribed about a given triangle, in terms of the sides of the triangle. Given the three angles of a plane triangle, and the radius of the inscribed circle, to find the sides of the triangle. Let A, B, C, be the three given angles, r the radius AB or c AP, + P,B C Given the three angles of a plane triangle, and the radius of the circumscribing As in Prob. 3. So, circle, to find the sides of the triangle. C B |