ANGULAR SURVEYING 71 



SOLUTION. By substituting the given values in the formulas, 



and 



DB 



47.50X (74.16- 50) 



22 - 95ch - 



Problem II. To cut off a given area by a line starting from 

 a given point on the boundary of a polygonal field. 



LetABCDEF, Fig. 2, be a field from which it is required to cut 

 off 5 acres by a line run through a given point G in the boundary. 

 Draw a line GD from G 

 to one of the opposite 

 angles of the plat in such a 

 position as to cut off an 

 area nearly equal to the 

 required area. Calculate 

 the length and bearing of 

 GD by the method given 

 under Supplying Omis- 

 sions. Calculate the area 

 GBCD, which will be called 

 Si. Find the difference 

 between the required area 

 5 and the calculated area Si. 

 additional area S f must be found; let GDH be this area. 

 Then, area GDH = S - Si = 5'. In the triangle GDH, the side 

 GD and the angle Z/ are known. Then, 



25' 



DH= 



GD sin ZX 



If the required area 5 is less than Si, the process is sub- 

 stantially the same, except that the required distance should 

 be calculated and measured from D along the line DC. 



EXAMPLE. In Fig. 2, assume that the length of the line 

 GD is 8.93 ch., that the angle GDH is 61, and that the area of 

 CBCD is 3.58 A. What must be the distance of the point 

 H from the point D, in order that the line GH will cut off 

 5 A.; that is, in order that the area of the figure GBCDH will 

 be 5 A.? 



E 



FIG. 2 

 If 5 is greater than Si, an 



