The angle in the tables corresponding to this logarithm, is 54° 2' 22", but we cannot determine a priori whether the angle sought be this angle, or its supplement 125° 57′ 38′′. Or B 54° 2' 22" B = 125° 57' 38" If we take the 1st value, C-84° 47' 38" and the triangle required is ABC If we take the second value, C=12° 52′ 22" and the triangle required is ADC = 145, A = see last fig. 41° 10', re This example also belongs to case 4, but since the given angle A is acute, and the side b opposite the required angle B less than the side a, it follows that the angle B must be an acute angle, and the solution will not be ambiguous. We have But log. sin. Blog. sin. A + log. blog. a = 9.8183919 log. b= 2.1613680 11.9797599 The angle in the tables corresponding to this logarithm is 32° 21' 54", and since, in the present instance, the supplement of 32° 21' 54" cannot belong to the case proposed, the solution is not ambiguous. Example 4. Given a 374, b = 3277.628, and the included. angle 57° 53' 16".8: required A, C, b. By case 3 we have log. tan. A-B 2 = C log. (ab)+log. cot.-log. (a+b) a-b-476.372, .. log. (a-b) 2 = 2.6779444 C log. cot. = 10.2572497 2 The angles A and B being determined, the side c may be readily found from the equation. C a sin. C log. c = log. a + log. sin. C-log. sin. A Example 5. Given a=33, b-42.6, c=53.6, required A, B, C. Taking the formula marked (s) in chap. III. we have log. sin. A =log. R+log. 2+}{log.s+log. (s—a)+log.(s—b)+log.(s—c)}—{log.blog.c} log. sin. B =log. R+log. 2+log. 8+log. (s—a)+log.(s—b)+log.(s—c)}-{log.a+log.c} log. sin. C =log. R+log. 2+1{log.s+log. (sa)+log.(s—b)+log.(s—c)}-{log.a+log.b} Now log.R=10. log.2=0.3010300 a=33 ..log.a =1.5185139 log.blog.c=3.3585744 b=42.6..log.b =1.6294096..log.a+log.c=3.2476787 c=53.6...log.c =1.7291648 log.a+log.b=3.1479235 s=64.6..log.s =1.8102325 s-a-31.6...log.s-a=1.4996871 s-b=22.log.s-b=1.3424227 s-c-11..log.s-c-1.0413927 ..log.s+log.(sa)+log.(s—b)+log.(s—c)=5.0937350 And {log.s+log.(s—a)+log.(s—b)+log.(s—c)} = 2.8468675 ..log.R+log.2+1{log.s+log.(s—a)+log.(s—b) +log.(s-c)} -= = 13.1478975 Subtracting from this number the values log. b+log. c; log. a+log. c; log. a+log. b; in succession we find, log. sin. A 9.7893231 .. A = 37° 59' 53' log. sin. B log. sin. C 9.9002188 .. B = 52° 37′ 46′′ = 9.9999740 .. C = 89° 22′ 20′′ Having determined A and B by the above method, we find the above accurate value of C, by subtracting the sum of A and B from 180°. If, however, it had been required to determine C alone (being an angle nearly equal to 90°) we could not have found its value with sufficient accuracy from the common tables, for it will be seen, upon referring to them, that the number 9.9999740 may be the logarithm of the sine of any angle from 89° 22" 20" up to 89° 22′ 25′′ consequently the above method cannot be applied with propriety to determine the exact value of C, unless we previously determine A and B. The angle C may however be determined directly, and with great accuracy, from any of the three formulæ (r), (3), (~), in Chap. III. Let us take these in succession, (♂). C log. cos.=log. R+ {log. s+log. (s—c)}—}(log.'a+ log. b) log.s=1.8102325).. {log.s+log.(s—c)}=1.4258126 log.(s-c)=1.0413927 2.8516252 log.a+log.b=3.1479235.. (log. a + log. b) log.R=10 11.4258126 = 1.5739617 log.R+{log.(c-a)+log.(s—b)} - {log.a+log.b} log.(s-a)=1.4996871). {log.(s-a)+log.(s-b)}= 1.4210549 By (8). log.sin. log.(s-a)=1.3424227 2.8421098 log.a+log.b=3.1479235 .. log. R=10. By (n). log. sin. = 9.8470932 C log.tan.log.R+{log.(s-a)+log.(s—b)}-{{log.s+log.(s—c)} log.(s-a)=1.4996871 log.(c-a)=1.3424227.. {log.(s-a)+log.(s-q)=1.4210549 2.8421098 log.s=1.8102325 log.s-c=1.0413927 2.8516252 CHAPTER VIII. ON THE USE OF SUBSIDIARY ANGLES. 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 the 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=a+b2-2ab cos. C. the side c is determined theoretically at once by this expression, but the formula 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. c2=a2+b2-2 ab cos. C Adding and substracting 2 ab on the right hand side. c2=a2+b2-2 ab+2 ab-2 ab cos. C =(a-b)2+2ab+2 sin. C c2=(ab) (1+tan.' ) c = (a—b) sec. log. c =log. (ab)+log. sec. -log. R The angle is known from the equation. C log.tan.q=log. 2+ † (log.a+log.b) + log. sin.—log.(a—b) 2 can be found from the ❤ being thus determined, log. sec. The angle o, 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 c2=a+b2-2 ab cos. C =a+b2+2ab-2 ab-2 ab cos. C =(a+b)2—2 ab×2 cos.' c2=(a+b)2 (1-sin.' ) =(a+b) cos.2 c=(a+b) cos. ❤ 2 log. c (s+b)+log. cos. -log. R As before the angle & must be determined from the equation. (a+c) in other words, that 2ab is always less than (a+b), this is easily done. |