68 ANGULAR SURVEYING 



SOLUTION. By calculation it is found that S* = 238.00 and 

 ^= 636.63. Taking G x as S 57 46' W and G y as N 79 47' W 

 and substituting in the formulas, 



__ 636.63 (-cos 57 46') -238 (-sin 57 46') _ 



y ~ (-sin 57 46') cos 79 47'- (-cos 57 46') (-sin 79 47') 



-339.56+201.32 



636.63+204.8 (-sin 79 47') 



3. When the bearings of two sides are missing, let 

 ,, 



Then, cos G y = 



from which G y is found, thus reducing the remainder of the 

 problem to case 1. 



EXAMPLE. The bearings and lengths of two sides of a field 

 are N 52 00' E, 10.63 ch., and S 29 45' E, 4.10 ch. The bear- 

 ings of the other two sides are to be determined, their lengths 

 being 7.69 ch. and 7.13 ch., respectively. 



SOLUTION. By calculation, S t = 6.54 -3.56 = 2.98 and S f 

 = 8.38+2.03 = 10.41. Then, 



7.132-7.692+2.982 + 10.412 



2X7.13 

 and 



-2.98 X 7.75 [- 10.4 1A/2.982- 7.752+ 10.41'] 



cos Gv = 



2.982+10.412 



= - .8682, or .47425 



Suppose that it can be seen from a sketch that the bearing G y 

 is northwest. Then the cosine will be positive, and the angle 

 corresponding to .47425 is the correct bearing; that is, G y 

 - N 61 41' W. Then, applying the method illustrated in 

 case 1, -10.41-7.13 (-sin614lQ 



-2.98-7.13 cos 61 41' 



-. 64937 = tan 33 



6.36 





