Prof. Twining on ti 
Let BA=1, (fig. 2,) be a force 
BC=a, andthier, applied - the angle 
ABC=A, with the first. Apply two 
new forces at the equal angle CBE,— 
namely, BE, BD; and let BE=BA; 
also let BC : BA:: BE=BA : BD=— 
Call the resultant of BA, BC y, and L 
that of BE, BD vy’. Then we have 
Res. (BA, BE) + Res. (BC, BD) = Res. (y, y’). But, since y : ‘y’ 
tia $1, the resultant of y and y’ is a 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, 
1 2 
Res. (BA, BE) + (;+4] = 7 But Res. (BA, BE) = 2cos. A. 
Whence y=(1+2a cos. Aa), 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=(1+a)?, 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 \ 
& 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 BI, having to BE the ratio of BA to AC. Com- 
plete EBGJ and join BJ. Complete CBJI,* and draw JH and 
ane 
* Since JH=FE=AB, and HJI, JHI equal ABC, BAC .-. BH passes through I. 
Fig, 3. 
B 
