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

Given  P+Q = 16N      , R= 8N

Angle between minimum force and R is 90⁰

$\begin{array}{l}\therefore Q=\sqrt{P^2+R^2}\\ \sqrt{64+{\left(16-Q\right)}^2}\end{array}$  $p=16-Q$

$Q=\sqrt{64+256+Q^2-32Q}$

$\cancel{Q^2}=320+\cancel{Q^2}-32Q$

$\begin{array}{l}\Rightarrow 32Q=320\\ \because P+Q=16\\ \because P=16-Q=16-10=6N\\ \because P=6N,Q=10N\end{array}$