The resultant of these forces OP, OQ, OR, OS and OT is approximately ...... N. [Take, $\sqrt{3}=1.7, \sqrt{2}=1.4$ and given $\hat{\mathrm{i}}$ and $\hat{\mathrm{j}}$ unit vectors along X, Y axis]

From given figure, we can write all the forces in vector notation as

           $\begin{array}{l}\mathrm{OP}=\left(20\sin {30}^{\circ }\hat{\mathrm{i}}+20\cos {30}^{\circ }\hat{\mathrm{j}}\right)\mathrm{N}\\ =(10\hat{\mathrm{i}}+17.32\hat{\mathrm{j}})\mathrm{N}\\ \mathrm{OQ}=\left(10\cos {30}^{\circ }\hat{\mathrm{i}}+10\sin {30}^{\circ }\hat{\mathrm{j}}\right)\mathrm{N}\\ =(8.66\hat{\mathrm{i}}+5\hat{\mathrm{j}})\mathrm{N}\\ \mathrm{OR}=\left(20\cos {45}^{\circ }\hat{\mathrm{i}}-20\sin {45}^{\circ }\hat{\mathrm{j}}\right)\mathrm{N}\\ =(14.14\hat{\mathrm{i}}-14.14\hat{\mathrm{j}})\mathrm{N}\\ \mathrm{OS}=\left(-15\cos {45}^{\circ }\hat{\mathrm{i}}-15\sin {45}^{\circ }\hat{\mathrm{j}}\right)\mathrm{N}\\ =(-10.60\hat{\mathrm{i}}-10.60\hat{\mathrm{j}})\mathrm{N}\\ \mathrm{OT}=\left(-15\sin {60}^{\circ }\hat{\mathrm{i}}+15\cos {60}^{\circ }\hat{\mathrm{j}}\right)\mathrm{N}\\ =(-12.99\hat{\mathrm{i}}+7.5\hat{\mathrm{j}})\mathrm{N}\end{array}$

The resultant of these forces is given as

           $\mathrm{F}=\mathrm{OP}+\mathrm{OQ}+\mathrm{OR}+\mathrm{OS}+\mathrm{OT}$           $$\begin{gathered} \mathrm{F}=\left[\right(10\hat{\mathrm{i}}+17.32\hat{\mathrm{j}})+(8.66\hat{\mathrm{i}}+5\hat{\mathrm{j}})+(14.14\hat{\mathrm{i}}-14.14\hat{\mathrm{i}})+(-10.60\hat{\mathrm{i}}-10.60\hat{\mathrm{j}})+(-12.99\hat{\mathrm{i}}+7.5\hat{\mathrm{j}})] \\ \mathrm{F}=9.21\hat{\mathrm{i}}+5.08\hat{\mathrm{j}} \mathrm{N} \\ \mathrm{F}\approx 9.25\hat{\mathrm{i}}+5\hat{\mathrm{j}} \mathrm{N} \end{gathered}$$

Thus, the resultant force is

       $\mathrm{F}\approx 9.25\hat{\mathrm{i}}+5\hat{\mathrm{j}} \mathrm{N}$