Four concurrent coplanar forces in newton are acting at a point and keep it in equilibrium (figure). Then values of F and $\mathrm{\theta }$ are

In equilibrium position
Along y-direction,
$2\sin {60}^{\circ }=\sqrt{3}+\mathrm{Fcos}\mathrm{\theta }$
or $2\times \frac{\sqrt{3}}{2}=\sqrt{3}+\mathrm{Fcos}\mathrm{\theta }$
or $\mathrm{Fcos}\mathrm{\theta }=0$
As $\mathrm{F}\ne 0$
$\therefore \cos \mathrm{\theta }=0\Rightarrow \mathrm{\theta }={90}^{\circ }$
Along x-direction
$\mathrm{Fsin}{90}^{\circ }=1+2\cos {60}^{\circ }=1+2\times \frac{1}{2}$
or F = 2 N