arith. co. log tang AB (38° 25') = 10.100692 AC + BC log tang (68° 22°4) = 10.401838 AC_BC 9.185174 AD-u log tang HO37701 2 log tang AD (64° 24') 10.319556 10. 9.680030 This logarithm corresponds to 61° 24'. Consequently the angle A=610 24. Yet we have elsewhere supposed it to be (99) 610 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 640 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 610 25'. 110. This last case may be resolved by one proportion, which A А B Q draw AL, 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 arc AC, and OP that of the arc AB. By (43) we have in general coc A_AQ’+APP-PQ or, 2 APX AQ whence PQ-=sec? b + sec ~2 sec 6 sec c cos a. 1 cos A= cos a sec.6 sec C-1 1 COS @ 1 cos 6 cos c tang ố tang 2 cos acos 6 cos c cos 6 Again, since cos 6 tang b=sin b, the expression becomes cos arcos 6 cos c cos A = sin b sin c Consequently I +cos A cos arcos 6 cos c—sin b sin c) sin b sin c cos acos (b+c) by (11) sin 6 cos c P+Q But in general (17) we have cos Q-cos P=2 sin 2 P-Q x sin 2 sinP7Q sin b sin c and therefore 1 +cos A= 2 sin S. sin (S-a) sin b sin c by taking s=a+b+c COS 2 2A But we have shewn that I +cos A=2 cos Therefore 2 and if radius=1 R’. sin S. sin (S-a) or, cos if radius=R sin b sin c Example. In the oblique-angled spherical triangle ABC, (preceding figure) given AB=76° 50'; AC=77° 5'; and BC=59° 40'. Required the angle A. By the formula we have A R’. sin S. sin (S—BC) But sin AB sin AC COS 76° 50' log R* = 20. 77° 5 log sin S (106° 47') = 9.981095 59° 40' log sin (S-BC) 9.864950 2213° 35 co. log sin AB (76° 50') = 10.011570 106° 47' =S co. log sin AC (77° 5') = 10.011131 59° 40' 2|19.868746 47° 7'=S--BC А log cos 9.934373 A Whence = 30° 42', and A=61° 24, the same as before. XII. 111. Given the three angles, to find a side, AC for example. This case resolved by the supplemental triangle would not differ from the preceding problem. But it may also be resolved by the two following proportions, if we A suppose that the arc CE bisects the an. gle ACB. For we have first Ε B A+B B-A cot : tang :: tang } C : tang DCB 2 2 and then, 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=610 25', B=82° 36', C=82° 4', we have A+B 2 B-A 2 9.939673 9.699941 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 ? arith. co. log tang A (61° 25') = 189.736269 log cot ACD (67° 38' 59'') 9.614006 10. 9.350275 Hence the side AC=77° 3' 18". We have before found it 77° 5', in the solution of the first case, and this difference arises from the quantities neglected in the valuation of the logarithms. 112. This case may also be solved a single proportion. For retaining the notation used in the last paragraph of the preceding case, let a', b, c', A', B, C', represent the sides and angles of the supplemental triangle ; and call a'+b+c=s. Then according to 2 the last problem A' sin S. sin (S'-a') 2 sin 6 sin c But by the definition of the supplemental triangle A 180—a a =sin 2 Similarly sin b=sin (180°-B)=sin B; and sin c=sin C. a' +'+c Again, sin S=sin sin 2 =sin (2700-S)=-cos S COS COS =COS 2 { and sin (S'-d')=sin{270.--S--(180---A)} = sin sin 900 =cos (S-A). Substitute these values in the above expression for cos. and we 2 if radius=l, or R?. cos S. cos (S-A), if radius=R sin B sin C The expression under the radical sign is always positive ; for ES has been shewn (70) to be greater than 90°, and less than 2709, and therefore cos S is negative, and consequently -cos S, positive. Again, s–A=B+C-4, but since in every spherical triangle the sum of any two sides is greater than the third side, we have sin sin A+B+C 2 2 b' +c'> a'; therefore 180°-B+180°-C> 180°_A, and consc B+C-A quently by transposition and reduction < 90°; whence 2 cos (S-A) is positive. Example. Given the three angles of an oblique spherical triangle respectively 61° 25', 82° 36', and 82° 4' ; required one of the sides; for example that opposite the angle 82' 36'. By the preceding formula we have side sought sin :--cos S. cos (S-82° 36') 2 sin 610 25' x sin 82° 4' 610 25' log R? 20. 820 36 log cos (1130 2') 9.592473 82° 4' log cos (30° 26') 9.935618 212260 5 arith. co. log sin (61° 25') = 10.056445 113° 2' –S arith. co. log sin (82° 4') = 10.004177 82° 36' 219.588713 300 26'-S-82° 36' side required 9.794356 2 Whence the side is 77° 3' nearly, as before. log sin EXAMPLES OF THE CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. AC = 1. In the oblique-angled spherical triangle ABC given BC=114° 30', 56° 40', and the angle BAC= 1250 20'; required the other parts of the B C triangle? Answer, 2 B=48° 30', 2 C=629 54, side AB=830 11. 2. In the oblique-angled spherical triangle ABC, given AB=45°30', BC=720 40ʻ, and the angle A=135o; required the rest ? Answer, L C=31° 53' 35", L B=24° 31 25", AC=34° 4' 46". 3. In the oblique-angled spherical triangle akc, given BC=1140 30', CA=56° 40', and the angle C=62° 54'; required the rest ? Answer, L B=730 31', LA=125° 20', side AB=830 12'. 4. In the oblique-angled spherical triangle ABC, given AB=41° 45, BC=730 20', and the angle B=30° 30'; required the rest ? Answer, 4 A=130° 8' 45", L C=31° 33' 43', AC=400 12' 59". |