First, (90) R: cos AC:: tang A cot ACD And then, (86) cos BCD: cos ACD:: tang AC: tang BC. VI. 104. Suppose any two sides given, AC and BC for example, with one of the angles opposite to those sides, such as the angle A; to find the angle B, opposite to the other side. This problem may be resolved by the well known proportion sin BC: sin AC:: sin A: sin B A If we assign to the given sides and angle A, the same values as in the preceding problems, we shall find the angle B by the following calculation: arith, co. log sin BC (59° 40′) = 10.063938 log sin AC (779 5') = 9.988869 = 9.943555 9.996362 The angle B is therefore 82° 36', as we have before supposed it, or 97° 24', as we might also have supposed it. This problem falls within the first case; they differ only by an inversion of the terms in the proportion which resolves both. VII. 105. The same things being given, required the third side AB. (87) R: cos A:: tang AC: tang AD (83) cos AC: cos BC:: cos AD: cos BD. By the first of these two proportions we find AD, and by the second BD: therefore we ac C A B B shall have AB= E VIII. 106. On the same suppositions, to find the angle C, contained between the given sides AC and BC. (90) Rcs:: tang A pot ACD, and, (88 tong BC: tang AC:s ACD: BCD. The supplemental triangle is equally proper to resolve this problem by reducing it to the third case. IX. 107. Given the two sides AC and AB, and the angle A, contained between them, to find the other side BC. (Preceding figure). (87) R: cos A:: tang AC: tang AD (83) cos AD: cos BD :: cos AC : cos BC. X. 108. The same things being given, to find one of the other two angles, the angle B, for example. (Preceding figure.) 109. If the three sides were given, how should we find one of the three angles, A for example? (Preceding figure). This first proportion will give us the segment AD, and the following one will give the angle A. (87) tang AC tang AD :: R: cos A. Resuming the values we have before supposed, we shall have 2 The half difference of the two segments is 25° 59′, w. th being added to the half base 38° 25′, gives 64° 24′ for the grefter segment AD. This premised, we have then arith, co. log tang AC (77° 5) = 189.360474 log tang AD (64° 24') = 10.319556 10. 9.680030 This logarithm corresponds to 61° 24'. Consequently the angle A=61° 24'. Yet we have elsewhere supposed it to be (99) 61° 25', whence arises this small difference? It proceeds from the quantities neglected in the valuation of the logarithms; for if we had continued the calculation as far as seconds, we should have found the half difference of the segments 25° 58′ 30′′, which added to 38° 25', would have given 64° 23′ 30′′, for the value of AD. But the logarithm of tang 64° 23' 30", is 10.319394. This logarithm written instead of 10.319556; gives for the result 9.679868, which answers to 61° 25'. 110. This last case may be resolved by one proportion, which may be investigated in the following manner, and which will serve to exemplify the use of the trigonometrical formulas as given in the foregoing pages. In the oblique-angled spherical triangle ABC, call the side BC a; AC, b; and the side AB, c. Imagine O to be the center of the sphere, and from the point A, B draw AP, AQ, tangents to the arcs AB, AC. Then by Euclid, def. 6, Book XI, the angle QAP is equal to the spherical angle A, and the angle at O is measured by the arc BC. Draw OQ and OP, then will OA be the secant of the are AC, and OP that of the arc AB. By (43) we have in general cos A—AQ2+AP2—PQ2 cos Atang b+tang2 c—PQ3 b 2 tang 6 tang c or, 2. AP X AQ whence PQ tang b+tang2 c-2 táng 6 tang e cos A. For similar reasons, PQ2=sec2 b+sec2 c~2 sec b sec c cos a. and subtracting the preceding equation from this latter one, we have 0=1+1+2 tang b tang c cos A-2 sec b sec c cos a, because sec2-tang' R=1). Whence by transposition, DD cos b Again, since cos b tang b=sin b, the expression becomes cos a-cos b cos c cos a—(cos b cos c-sin b sin c) sin b cos c by (11) But in general (17) we have cos Q-cos P-2 sin 2A But we have shewn that 1+cos A=2 cos. Therefore In the oblique-angled spherical triangle ABC, (preceding figure) given AB=76° 50'; AC-77° 5' ; and BC 59° 40'. Required the angle A. Whence 30° 42′, and A=61° 24, the same as before. tang A cot ACD: ; R: cos AC. The first proportion gives the value of the angle DCE, which being added to the angle ACE, will shew the angle ACD. And this once known, we directly obtain the required side AC, by the second proportion. Let then A=61° 25′, B=82° 36′, C=82° 4′, we have arith. co. log cot + (72° 0′ 30′′) = 2 10.488439 = 9.939673 Therefore the angle formed by the perpendicular arc, and by that which bisects the angle C, must in this case be 26° 36′ 59′′. Adding this value to the half of the angle C, we shall have 67° 38′ 59′′, for the value of the angle ACD; and the second proportion gives |