Prof. Twining on the Parallelogram of Forces. 327 



Let BA=1, (fig. 2,) be a force, and 

 BC = a, another, applied at the angle 

 ABC=A, with the first. Apply two 

 new forces at the equal angle CBE, — 

 namely, BE, BD; and let BE=BA; 



1 



also let BC : BA: :BE=BA : BD=- 



a 



Call the resultant of BA, BC y, and 



that of BE, BD y'. Then we have 



Res. (BA, BE) -f Res. (BC, BD) = Res. (y, y'). 



r 



But, since y : y' 



::«:!, the resultant of y and y' is — ; and it must lie in the 



line BD, because it must make the same angle with the force y' 

 that the resultant y' makes with the force BD. We have, then, 



Res. (BA, BE) -f I -+a) = "—• But Res. (BA, BE) = 2cos. A. 



Whence y=(l-f2acos. A+a a ) , which, by trigonometry, equals 

 the third side of a triangle having two sides AB and BC, and 

 their included angle the supplement of A. If we suppose A to 



be a right angle, our expression reduces to y=(l+a 2 y, and be- 

 comes an independent demonstration of the law that regulates 

 the intensity of a resultant compared with that of its rectangular 

 components. 



The necessity of the reference to 

 trigonometry, made above, may be 

 obviated by an application, purely ge- 

 ometrical, of the same principle of rea- 

 soning, in the following problem. 



Prob. To find one side of a triangle 

 from the other two sides which include 

 a given angle. 



Let BA, AC (fig. 3) be two given 

 sides which include a given angle BAC. 

 Draw BI parallel to AC ; also BE equal 

 to BA, at the angle EBI equal to ABI. 

 Take BG in BT, having to BE the ratio of BA to AC. Com- 

 plete EBGJ and join BJ. Complete CBJI * and draw JH and 

 EF parallel to AB. 



* Since JH=FE=AB, and HJI, JHI equal ABC, BAC .-. BH passes through I. 



