126 



HYDROGRAPHIC SURVEYING 



After X and F are found the distances FE and FD can be 

 figured by trigonometry. 



EXAMPLE. Given a = 850 ft., 6 = 760 ft., W=150. A = 41 

 30', and 5 = 35 30'. What are the values of EF and DFJ 



SOLUTION. Substituting the given values, 5 = 360 227 

 = 133 and cot 5= -cot (180 -S)= -cot 47. 

 Substituting known values in the preceding formula, 



850 sin 35 30> 

 001 * 



X = 67 49'; whence, F-=133-67 49' = 65 II 7 . 



In the triangle FCE. CF=180-(41 30 / +67 49') = 70 



no 



sin 41 "SO' 



In the triangle DCF, DCF=18Q-(35 t 

 79 19'. Therefore, 



