30 



CHAIN SURVEYING 



as C, and the end of the ninetieth link at the other extremity B. 

 Hold the end of the fiftieth link and draw the chain until both 

 parts are taut. The point D where the end of the fiftieth link 

 is held will then be a point in the perpendicular, and the direc- 

 tion of the latter will therefore be BD. 



The distance BC may be any other convenient multiple of 3. 

 In general, if BC is denoted by 3 a, BD must be 4 a, and CD 

 must be 5 a. Thus, BC may be made equal to 21 ( = 3X7) li.; 

 in which case BD must be 4X7 = 28, and CD must be 5X7 

 = 35, li. As 35+28 = 63, one end of the chain must be fixed 

 at one of the extremities of BC, the end of the sixty-third link 

 at the other extremity, and the chain pulled from the end of 

 the thirty-fifth link until both parts are taut. 



To Determine the Angle Between Two Lines. Let AD and 



AE, Fig. 3, be two lines on the ground. To deterrr.ine the 

 angle DAE, measure off from A on AD and AE equal dis- 

 tances AB and AC. Measure the distance BC. Then the 

 angle DAE is calculated from the relation 



\BC 

 sin \DAR = - 



EXAMPLE. If AB and AC are each 100 ft. and BC is 57.6 ft., 

 what is the value of the angle DAE? 



SOLUTION. Substituting the values of BC and AB in the 

 preceding equation, 





= . 28800; 



whence, \DAE 

 X2 = 3328'. 



16 44', nearly; and, therefore, DAE= 16 44' 



