ANGULAR SURVEYING 



73 



Problem m. To determine the length of a line both ends of 

 which are inaccessible. 



Let AB, Fig. 5, be the line, the ends A and B of which are 

 inaccessible. Select two points P, Q from which both ends of 

 the line can be seen, and at a dis- 

 tance from each other of about 300 

 or 400 ft. Measure the line PQ, 

 and the angles K, L, M, and N. 

 Then, from triangle APQ, 

 PQ sin M 



AP= ; 

 sin R 



in which R = 180 - (K+L) - M. 

 From triangle BPQ, 



, PQ sin (M+N) 



>P = ~ - " - ' 



sin 5 



in which S = 180 L-(M + N). 

 Then, from triangle A BP, 



FIG. 5 





Finally , 



EXAMPLE. If, in Fig. 5, the distance PQ is 400 ft., and the 

 angles, as measured, are K = 37 10',Z. = 36 30', M = 52 15', 

 N = 32 55', what is the distance ABf 



SOLUTION. In the triangle APQ, R = 180- (37 iy+36 30 7 

 +52 150 = 54 05', and 



400 sin 52 15> 



AP = =390.53 ft. 



sin 54 05' 



In the triangle BPQ, S=180-(36 30 / +52 15'+32 55') 

 = 58 20', M+N = 52 15'+32 55' = 85 10', and 



Also, -K = 37 10', } K= 18 35', and 



whence, 



A3 



468.30+390.53 



(X- Y) = 15 04', and therefore 

 (468.30-390.53) cos 18 35' 



sin 15 04' 



283.58 ft, 



