New plane and spherical trigonometry, surveying, and navigation (1903)
Wentworth, G. A. (George Albert), 1835-1906 · Boston : Ginn & Company · leaf 268
ight triangle J.^/C,sin \ a — sin a sin -£- J..sin ia sin £ J. =By [12], = sin asin i a 2 sin £ a cos £ a= -J sec | a. 3. Given the side a of a regularspherical polygon of n sides; findthe angle A of the polygon, thedistance R from the centre of thepolygon to one of the vertices, andthe distance r… … Then ACE = ACM = 360° > n 180° CAM = I A,AM = i a,AC = R,MC = r. By Napiers Rules, 180° 1 • 1 A cos = cos $ a sin ■$• A, n • r, • 180° sin i a = sm R sin , n sin r = tan -J- a cot 180° Whence, 180°smii = sec i a cos > n sin R = sin £ a escsin r == tan i a cot 180° n180° n 260 SPHERICAL…
Trigonometry · Surveying · Navigation
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