The sum of magnitude of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to minimum force then the forces are

A+B = 16(given) ……..(i)

$\mathrm{tan\alpha } = \frac{\mathrm{Bsin\theta }}{\mathrm{A}+\mathrm{Bcos\theta }} = \tan {90}^0$

$\therefore$A+$\mathrm{Bcos\theta } = 0 \Rightarrow \mathrm{cos\theta } = \frac{-\mathrm{A}}{\mathrm{B}}$ ……(ii)

8 = $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos} \mathrm{\theta }}$  ……..(iii)

By solving eqs. (i), (ii) and (iii) we get

A = 6 N, B = 10 N