ANGULAR SURVEYING 67 



EXAMPLE. The bearings and lengths of the first three 

 courses of a survey are, respectively, N 32 15' E, 22 ch.; 

 S 36 30' E, 10 ch.; and S 15 45' E, 5 ch. Determine 

 the length and bearing of the fourth course, which closes the 

 survey. 



SOLUTION. Let gi, gz, and e be the longitude ranges, and 

 /i, /2, and ta the latitude ranges of the known courses, which 

 are as follows: 



gi = 22 sin 32 15' = 22X.53361= 11.74 

 g2 = 10 sin 36 30' = 10X.59482 = 5.95 

 ga= 5 sin 15 45'= 5X.27144 = 1.36 



19.05 ch. = 3g- 



*i= 22 cos 32 15'= 22X. 84573= 18.61 

 h = - 10 cos 36 30' = - 10 X .80386 = - 8.04 

 / 3 = -5 cos 15 45'=- 5X. 96246 =-4.81 



+ 5.76 = 5, 

 Then, g x = 19.05 and i x = 5.76. Therefore, tan G x 



= -; whence, G x = 73 ll'. The bearing is S 73 11' W. 



5.76 

 Also, as both ranges are negative, 



2. When the lengths of two sides are missing, let l x and l y be 

 these lengths of the deficient sides, and G x and G y their corre- 

 sponding bearings. Then, 



_ Sg- cos G x S t sin G x 

 V sin G x cos G y cos G x sin G y 



Sgr+ly sin G y 



and l x = -- : 



sin G x 



EXAMPLE. In a six-sided field, the lengths and bearings 

 of four sides are N 30 36' E, 314 ft.; N 89 35' E, 406.0'ft.; 

 S 32 14' E, 212.0 ft.; and N 26 15' W, 196.2 ft. The bear- 

 ings of the other two sides are S 57 46' W and N 79 47' W. 

 Determine their lengths. 



