New plane and spherical trigonometry, surveying, and navigation (1903)
Wentworth, G. A. (George Albert), 1835-1906 · Boston : Ginn & Company · leaf 147
A the position of the balloon,and A the point directly under A,in the same horizontal plane as BC.Then AA -= AB tan ABAsinACB^ BC- = 3000 sin BACsin 64° 30 tan ABA tan 18° sin 40° 5 log 3000 = 3.47712 log sin 64° 30= 9.95549 log tan 18° = 9.51178 colog sin 40° 5 = 0.19118 log A A= 3.13557 AA =… … Let AB and AD represent theforces, and AG their resultant.Then, in the triangle ABC, givenc = 410, a = 320, B = 180° - 51° 37= 128° 23; required b and A. tan |(C - A)c — a c -f a90 tani(0-f A) - tan 25° 48 30.730 log 90 = 1.95424 log tan 25° 48 30 = 9.68448 colog 730 = 7.13668-10 tan-i(C-^i) =…
Trigonometry · Surveying · Navigation
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