Problem 361 | Equilibrium of Non-Concurrent Force System

From the Force Polygon
$\dfrac{R_A}{\sin 45^\circ} = \dfrac{20 + 30}{\sin 105^\circ}$

$R_A = 36.60 \, \text{ kN}$
 

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

$(4 \cos \theta)(R_A \cos 30^\circ) = (4 \sin \theta)(R_A \sin 30^\circ) + (3 \cos \theta)(20) + (1 \cos \theta)(30)$

$(4 \cos \theta)(36.60 \cos 30^\circ) = (4 \sin \theta)(36.60 \sin 30^\circ) + (3 \cos \theta)(20) + (1 \cos \theta)(30)$

$126.7861 \cos \theta = 73.2 \sin \theta + 60 \cos \theta + 30 \cos \theta$

$36.7861 \cos \theta = 73.2 \sin \theta$

$\dfrac{36.7861}{73.2} = \dfrac{\sin \theta}{\cos \theta}$

$\tan \theta = 0.502\,542\,349\,7$

$\theta = 26.68^\circ$           answer