The sum of the magnitudes of two forces acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at 90° with the force of smaller magnitude, what are the magnitudes of forces?

The correct option is C $5,13$
Let the force be A & B and A < B
$A+B=18\Rightarrow B=18-A$
$|\overrightarrow{A}+\overrightarrow{B}|=12$
$\sqrt{A^2+B^2+2AB\cos \theta }=12$
Also, $tan{90}^{\circ }=\frac{B\sin \theta }{A+B\cos \theta }$
$\Rightarrow A+B\cos \theta =0\Rightarrow \cos \theta =\frac{-A}{B}$
$\sqrt{A^2+B^2+2AB\left(\frac{-A}{B}\right)}=12$
$B^2-A^2=144$
$(18-A{)}^2-A^2=144$
$324-144=36A$
$A=5$
$B=18-A\Rightarrow B=13$