VOL. LXV.] . PHILOSOPHICAL TRANSACTIONS. 693 



A — az — A X VS. ^ = AB, whence z = ^^^^ _ ^ x vs. ^ = tang. « X co. 

 ver. sin. p — t. » X vs. ^. 



Ex.— Let A = 23° 28' 15", and ac = 80° C. 



,. Tang. A 9-6377 .lii mr, •> 



3'". Ve« sin. CO. AC, 10° ...8.I8I6 /"'"«' 



Sum, C nearly 22° 41' sine 7.8193 



Correction — I 



Gives C , ... 22 40, and BC = 23" 5' 35". 



This problem will be of use to find the declination of the ecliptic, and the 

 latitude of a planet near the limits. And these 3 instances will suffice for an 

 application of this method to right-angled spherical triangles ; we shall now give 

 2 problems of oblique triangles. 



Prob. 4. Suppose ABC to be a spherical triangle, in which are given the two 

 sides AB, BC, with the included angle b, tojind the third side ac. Fig. Q. 



Solut. 1. Let ABC = (3, BC = a, AB = <?. Put ac = (3 -f ^, j3 being an ap- 

 proximate value of c, when the two legs are nearly quadrants. Now the co- 

 sine of AC being equal to 6da -f- da,* we shall have b — bz — (i X vs) ; ^ = 



bvA + da : and z = -^ — -— X vs. ^. But i — da — da = vs. (^—a), 



which put = w. Then z ^ 1 ^^^ — x vs. ^ = cot. (3 X vs. (d—a,) 



— cos. J X COS. X X tang, -j- (3 — cot. |3 X vs. ^. Therefore i^ is the difference of 

 two arcs whose sines are cot. (3 X vs. (i — a), and cos. S X cos, a X tang. -fjS, the 

 difference of these two arcs being diminished by the correction cot. (3 X vs. ^. V 



Ex.— Suppose B = 5 1° 12' 5" t W 



AS = 87 57 51 '"-* 



BC = 87 20 34 



Cotangent b 9-9053 Tang. J b 25° 36' 9.6804 



Vers, sine AB — BC 00 37' 17'.. 5. 7693 Cosine a b 8.5506 -^ 



Cosine bc 8.6661 j 



Sum = 1st arc 0' 10'' sine 5.6746 2d arc 2' 43" sine 6.8971 



The difference of these two arcs, 2' 33" '^ 



Subtracted from the value of the angle b,. 51 12 5 



Leaves AC, 51 9 32 



The correction cot. /3 x vs. ^ in this example is 0. 



This solution is very convenient to find the distance of two zodiacal stars, 

 having their latitudes and difference of longitude. 



Solut. 1. Let T be an arc whose cosine t = b X cos. S — a,=: bda -f- ^da, and 

 suppose AC = T — ^, then < -|- tz — i X vs. ^ = buK -\- da =^ t ~ bda -j- da." 



* It is a well known theor. that sin. b a x sin. bc : r^ = vs. AC — vs. (ab — bc) : vs. b ; that is, 

 sin. BA X sin. bc ; r* = cos. (ab — bc) — cos. ac : r — cos. b. Or, in the author's notation, put- 

 ting r = I, DA : I = cos. (^— «) — cos. AC : I — J. Therefore da — Ada = cos. {i — ») — cos. 

 AC. Or, cos. AC = 6da — da — cos. (^— «.) For cos. (i - ») substitute its value as expressed 

 in the second corollary of the lemma, and there arises the author's equation, cos. ac = bo a + da. 



Orig. S. HORSLET. 



