There are two forces each of magnitude 10 units. One inclined at an angle of 30° and the other at an angle of 135° to the positive direction of x-axis. The x and y components of the resultant are respectively,

Here, $A-OP=10unitsalongOP$

$B-\left(OQ\right)=10unitsalongOQ$

$\therefore \angle XOP={30}^0$

and $\therefore \angle XOQ={135}^0$

$\therefore \angle QOX\text{'}={180}^0-{135}^0={45}^0$ 

Resolving A and B into two rectangular components we have A cos30° along OX and A sin30° along OY. While B cos 45° along OX’ and B sin 45° along OY’. Resultant of components of forces along x-axis.

$=(A\cos {30}^0-B\cos {45}^0)\widehat{i}$

$=(10\times \sqrt{3}/2-10\times 1\sqrt{2})\widehat{i}=1.59\widehat{i}$

Resultant of components forces along y-axis

$=(A\sin {30}^0+B\sin {45}^0)\widehat{j}$

$=(10\times \frac{1}{2}+10\times \frac{1}{\sqrt{2}})\widehat{j}=12.07\widehat{j}$