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

$\overrightarrow{\mathrm{F}}=\overrightarrow{{\mathrm{F}}_1}+\overrightarrow{{\mathrm{F}}_2}\text{; suppose that }{\mathrm{F}}_1<{\mathrm{F}}_2$

$\text{ given }{\mathrm{F}}_1+{\mathrm{F}}_2=16\to \left(1\right)\mathrm{F}=8$

$64={\mathrm{F}}_1^2+{\mathrm{F}}_2^2+2{\mathrm{F}}_1{\mathrm{F}}_2\cos \theta \to \left(2\right)\text{ and }\alpha ={90}^{\circ }$

$\Rightarrow \mathrm{Tan}\alpha =\frac{{\mathrm{F}}_2\sin \theta }{{\mathrm{F}}_1+{\mathrm{F}}_2\cos \theta }$

$\Rightarrow F_1+F_2\cos \theta =0\to \left(3\right);\text{ From (2) and (3) }$

$64={\mathrm{F}}_1^2+{\mathrm{F}}_2^2-2{\mathrm{F}}_1^2;64={\mathrm{F}}_2^2-{\mathrm{F}}_1^2\left[\text{ from (1) }\right]$

$\Rightarrow {\mathrm{F}}_2-{\mathrm{F}}_1=4\to \left(4\right)$

$\text{ from (1) and (4) }{F}_1=6N,F_2=10\mathrm{N}$