Five forces 2 N, $\sqrt{3}\mathrm{N}, 5 \mathrm{N}, \sqrt{3 } \mathrm{and} 2 \mathrm{N} \mathrm{respectively} \mathrm{act} \mathrm{at} \mathrm{a} \mathrm{particle} \mathrm{P} \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{figure}.$

The resultant force on the particle P is

$\sum _{ }{\mathrm{F}}_{\mathrm{x}} = \mathrm{sum} \mathrm{of} \mathrm{x}-\mathrm{component} \mathrm{of} \mathrm{forces}$

             = $2+\sqrt{3} \cos {30}^0+5 \cos {60}^0-2 \sin {30}^0$

               = $2+\frac{3}{2}+\frac{5}{2}-1 = 5 \mathrm{N}$

$\sum _{ }{\mathrm{F}}_{\mathrm{y}} = \mathrm{sum} \mathrm{of} \mathrm{y}-\mathrm{component} \mathrm{of} \mathrm{forces}$

            =  $\sqrt{3} \sin {30}^0+5 \sin {60}^0 + \sqrt{3}+2 \cos {30}^0$

            = $\frac{\sqrt{3}}{2}+\frac{5\sqrt{3}}{2}+\sqrt{3}+\sqrt{3} = 5\sqrt{3} \mathrm{N}$

$\therefore \text{\hspace{0.17em}F = }\sqrt{{\left(\sum _{ }{\mathrm{F}}_{\mathrm{x}}\right)}^2+{\left(\sum _{ }{\mathrm{F}}_{\mathrm{y}}\right)}^2}\text{ = }\sqrt{5^2+{\left(5\sqrt{3}\right)}^2}$

         = $\sqrt{100} = 10 \mathrm{N}$

$\therefore \mathrm{tan\theta } = \frac{{\displaystyle \sum _{ }{\mathrm{F}}_{\mathrm{y}}}}{{\displaystyle \underset{ }{\sum {\mathrm{F}}_{\mathrm{x}}}}} = \frac{5\sqrt{3}}{5} = \sqrt{3} \Rightarrow \mathrm{\theta } = {60}^0$