New plane and spherical trigonometry, surveying, and navigation (1903)
Wentworth, G. A. (George Albert), 1835-1906 · Boston : Ginn & Company · leaf 152
jects. Let A be the top of the hill, Athe point directly under A in thehorizontal plane of the base of thehill, B and C the objects observed. Then AB = AA. AC = AA tan 60° = VsAA. BC2 = 2FJB2 + WC1 - 2 AB x AC cos BAC= A7A2 + 3A7A2 -2 AA x V3 AA x i V3= AA2 + 3Z7Z2 - 3 ATA2 = AA\BC = AA. 62.… … tan ADC = AC 100 CD 200 2log tan ADC = 9.69897. ADC = 26° 33 54.5DC = ^1D5 + -4£>C= 63° 52 24.5C = ODtan5Z)C = 200 tan 63° 52 24.log 200= 2.30103log tan 63° 52 24 = 10.30939log BC = 2.610425(7 = 407.77.^LB = 5C-^iC = 307.77..45 = 307.77 yards. 63. The sum of the sides of atriangle is 100. The…
Trigonometry · Surveying · Navigation
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