Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F 1 to F 2 is 1 :…

Here, $\theta =45°$

Applying equilibrium conditions,

Taking components along $x$ & $\mathrm{y}$

Considering right side positive, $\sum F_x=0$

$$\begin{gathered} F_1+1\sin 45°-2\sin 45°=0 \\ {F}_1=2\sin 45°-1\sin 45° \end{gathered}$$

$F_1=\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{2-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

Taking upward positive,$\sum F_y=0$

$$\begin{gathered} 1\cos 45°+2\cos 45°-F_2=0 \\ {F}_2=1\cos 45°+2\cos 45° \end{gathered}$$

$F_2=\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

$F_1:F_2=1:3$

$x=3$