Four forces are acting at a point P in equilibrium as shown in...

Here, $\theta = 45 °$

Applying equilibrium conditions,

Taking components along $x$ & $y$

Considering right side positive, $∑ {F}_{x} = 0$

${F}_{1} + 1\sin 45 ° - 2\sin 45 ° = 0{F}_{1} = 2\sin 45 ° - 1\sin 45 °$

${F}_{1} = \sqrt{2} - \dfrac{1}{\sqrt{2}} = \dfrac{2 - 1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

Taking upward positive,$∑ {F}_{y} = 0$

$1\cos 45 ° + 2\cos 45 ° - {F}_{2} = 0{F}_{2} = 1\cos 45 ° + 2\cos 45 °$

${F}_{2} = \sqrt{2} + \dfrac{1}{\sqrt{2}} = \dfrac{2 + 1}{\sqrt{2}} = \dfrac{3}{\sqrt{2}}$

${F}_{1}: {F}_{2} = 1: 3$

$x = 3$