Problem 359 | Equilibrium of Non-Concurrent Force System

From the Force Polygon
$\dfrac{R_A}{\sin 45^\circ} = \dfrac{20 + 10}{\sin 105^\circ}$

$R_A = 21.96 \, \text{ kN}$
 

From the Free Body Diagram
$\Sigma M_B = 0$

$4(R_A \cos 30^\circ) = 20(3) + 10x$

$4(21.96 \cos 30^\circ) = 20(3) + 10x$

$10x = 16.072$

$x = 1.61 \, \text{ m}$           answer