ANGULAR SURVEYING 69 



As both ranges are negative, the bearing is S 33 W. 

 4. When the length l x of one side and the bearing G y of 

 another are missing, l x is determined by the formula: 



l x = Sg sin G x S( cos G x 



S t cos G X Y 



When l x has been determined, the unknown bearing G y is 

 found as in case 1. 



NOTE. In the two preceding cases, two sets of results will 

 generally be obtained. The problems are therefore indeter- 

 minate. However, if the notes contain a sketch showing the 

 shape of the tract, both sets may be plotted and the correct 

 figure identified. 



EXAMPLE. Two sides of a four-sided field have the bear- 

 ings and lengths N 77 24' W, 32 ch., and N 38 49' E, 14 ch. 

 The other two sides are deficient, one having the length 32.52 

 ch., bearing unknown, and the other the bearing S 18 15' W, 

 length unknown. 



SOLUTION. In this example the required values are l x and 

 G y . By calculation S ff = 22.45 and 5^=17.89. Then, sub- 

 stituting known values in the formula, /* = 28.2 ch. or 8.3 ch. 



As the second value of l x is negative, it shows that in this 

 case only one solution is possible. 



The required bearing G y is now determined as in case 1. 

 Thus ^ = 22.45+28.2 sin 1815' = 31.28, and t y = -17.89+28.2 



OJ OO 



cos 1815' = 8.89. Then, tan G y = = tan 7408', and the 



8.89 



bearing is N 7408' E. 



In applying these formulas, careful attention should be given 

 to the algebraic signs of the functions and of the ranges, which 

 signs depend on the bearings. For northeast and northwest 

 bearings, the latitude ranges and the cosines are +, and for 

 southeast and southwest bearings, the cosines are ; the 

 longitude ranges and the sines are + for northeast and south- 

 east bearings, and for northwest and southwest bearings. 



PROBLEMS ON DIVISION OF LAND 



Problem I. To divide a trapezoid into two parts, whose areas 

 shall be proportional to two given numbers, by a line parallel to the 

 bases. 



