692 PHILOSOPHICAL TRANSACTIONS. [aNNO 1775. 



of an arc, which arc we know is nearly equal to a. whose cosine is a. Suppose 

 the arc to be « — (3, and its cosine a-^ c. Then a-^ c= cos. (a — |3) = a 

 + AB — a X vs. p. Therefore, b = — + - X vers. (3. The first term - 



A A A 



will always give a near approximation to the value of sin. (3 ; and |3 being found, 

 the correction, - X vs. |3, or cot. a X vs. (3, may be found and added to it. 

 Among the tables requisite to be used with the Nautical Almanack, is table 4 

 for parallax, p. IQ, which shows the value at sight of such quantities as vs. |3 

 X cot. a, the arc |3 being found in the first column of the table, and « at the 

 top. This table I have calculated only to arcs under 63' ; but it would be 

 found useful to have a table ready computed for all arcs under 5°. 



Prob. 1. — If the two legs, ab and bc, of the spherical triangle abc right-angled 

 at b, are given, to find the hypothenuse AC, the leg bc, being small in comparison 

 of AC. Fig. 8, pi. 13. — Let ab = a, bc = (3, and suppose ac = a + C> "■ 

 being a near approximation to ac, and ^ the small arc to be added to ab to 

 make it equal to ac ; then cos, ac = cos. ab X cos. bc ; that is, according to 

 our notation, a — az — (a X vs. Q = ab. Whence z = tnJL (f ^ vs.Q 



= cot. « X vs. (3 — cot. a. X vs. ^. 



Ex.— Let A B be 75° 0', and bc 20" C, then the computation will be as follows : 



Cotangent ab ,... 9.4280 



Versed sine bc 8.7804 



Sum C nearly 55' 33" sine 8.2084 



Correction — 7 from tab. 4 Nautical Almanack. ,Q 



Therefore ^ = 55 26, and ac = 75" 55' 26". 



By this problem, the distance of the sun may be found from a planet whose 

 latitude and difference of longitude are known. 



Prob. 1. — Having the hypothenuse, ac, and one of the angles a, to find the 

 base ab. — Let ac = (3, bac = a, fig. 8, and suppose ab = |3 — ^, then cos. 



... * . . B Z , B X vs. ? , Z , B X vs. ? 



A = cot. AC X tang, ab, or a = - x ^ - ^. + — p— = ^ - 7^ + -jT— • 



Whence z = fii X (1 — c) + | X vs. ^ = ^ 2 (3 X vs. a + tang. (3 X vs. ^. 



Ex.— Let A = 23° 28' 15", and ac = 10° 0' 0". 



Sine2Ac20°0' 9.5340 



Versed sine a 8.9177 



Sum 8.45 17 



Log. 2 0.3010 



Dif. Z, nearly 48' 39" sine 8.1507 



Correction + 6 



^ 48 45, and bc = 159" 11' 15". 



By this problem, the right ascension of any point of the ecliptic, whose 

 obliquity and longitude are known, may be found. 



Prob. 3. Supposing the same things known as in the last, to find the perpen- 

 dicular bc, when the hypothenuse is nearly a quadrant. — ^Let a = a, AC = (3, 

 fig. 8, as before, and suppose bc = a — ^; then sin. bc = sin. ac X sin. a, or 



