The force polygon below is isosceles, thus, $R_B = W$
$\Sigma M_A = 0$
$(R_B \cos 30^\circ)(2L \cos \alpha) = W(L \cos \alpha) + (R_B \sin 30^\circ)(2L \sin \alpha)$
$(W \cos 30^\circ)(2L \cos \alpha) = W(L \cos \alpha) + (W \sin 30^\circ)(2L \sin \alpha)$
$2WL \cos 30^\circ \cos \alpha = WL \cos \alpha + 2WL \sin 30^\circ \sin \alpha$
$2 \cos 30^\circ \cos \alpha = \cos \alpha + 2 \sin 30^\circ \sin \alpha$
$2 \cos 30^\circ \cos \alpha - \cos \alpha = 2 \sin 30^\circ \sin \alpha$
$(2 \cos 30^\circ - 1) \cos \alpha = 2 \sin 30^\circ \sin \alpha$
$\dfrac{2 \cos 30^\circ - 1}{2 \sin 30^\circ} = \dfrac{\sin \alpha}{\cos \alpha}$
$\tan \alpha = \sqrt{3} - 1$
$\alpha = \arctan (\sqrt{3} - 1)$
$\alpha = 36.21^\circ$ answer
