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