## In This post you will get answer of Q.

Two forces are such that the sum of their magnitudes is $18 N$ and their resultant which has magnitude $12 N$, is perpendicular to the smaller force. Then the magnitudes of the forces are :

Solution:

## $A+B=18$…(1)

$ 12 =sqrt{A^{2}+B^{2}+2 A B cos theta}$…(2)

$tan alpha =frac{B sin theta}{A+B cos 0}$

$Rightarrow tan 90^{circ} =frac{B sin theta}{A+B cos theta}$

$Rightarrow cos theta=frac{-A}{B}$…(3)

Solving Eqs. (1), (2) and (3), $A=5 N, B=13 N$