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New plane and spherical trigonometry, surveying, and navigation (1903)
Wentworth, G. A. (George Albert), 1835-1906 · Boston : Ginn & Company · leaf 303
2 area J.F.E :2 area #DjE;JL2 area jBDC 6.4 x 9.25 = 59.2 : 13.75(4.75+ 7) = 161.5625: 10 x 4 = 40. 2 area ABCDEFarea ABCBEF130.38125 sq. ch. = 13.038125 a. = 260.7625= 130.38125= 13a. 6T\?< 7. Determine the area of the field ABCB from two interior stations,P and P, if PP = 1.50 chains, … PPC = 89° 35,PPB = 185° 30,PPOl = 309° 15,PPfB = 349° 45, PPB = 3° 35,PPA = 113° 45,PPB = 165° 40,PPC = 303° 15. Area = A P.4D + A PCB + A PBC + A P^i£. ZPPD= 10° 15,Z PDP = 4° 5,Z PPJS = 174° 30, Z PP^i = 50° 45,ZP^4P = 15° 30,Z PBP = 1° 55, Z PPC = 56° 45,Z PCP = 33° 40. TEACHERS EDITION. 295 PD…Trigonometry · Surveying · Navigation
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