Three forces $F_1,F_{2}and{F}_3$ are acting on particle P (shown in figure). If particles P is  in equilibrium, then 

 

According to Lami’s theorem 

$\frac{{\mathrm{F}}_1}{\sin (\mathrm{\pi }-\mathrm{\theta })}=\frac{{\mathrm{F}}_2}{\sin {135}^0}=\frac{{\mathrm{F}}_3}{\sin ({45}^0+\mathrm{\theta })}$ 

Or $\frac{4}{\mathrm{sin\theta }}=\frac{{\mathrm{F}}_2}{\frac{1}{\sqrt{2}}}=\frac{{\mathrm{F}}_3}{\sin {45}^0\mathrm{cos\theta }+\mathrm{sin\theta sin}{45}^0}$

$\text{or }\frac{\mathrm{4}}{\mathrm{sin\theta }}\mathrm{=}\sqrt{\mathrm{2}}{\mathrm{F}}_{\mathrm{2}}\mathrm{=}\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\mathrm{cos\theta }+\mathrm{sin\theta }}$

$\mathrm{or} 4\mathrm{cos\theta }=(10\sqrt{2}-4)\mathrm{sin\theta }$

$\Rightarrow \mathrm{tan\theta }=\frac{4}{10\sqrt{2-4}}=\frac{2}{5\sqrt{2}-2}$

$\Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{2}{5\sqrt{2-2}}\right)$

$\frac{\mathrm{4}}{-\mathrm{sin\theta }}\mathrm{=}\sqrt{\mathrm{2}}{\mathrm{F}}_{\mathrm{2}}\Rightarrow {\mathrm{F}}_{\mathrm{2}}\mathrm{sin\theta }=2\sqrt{\mathrm{2}}\mathrm{N}$