Let A and B be the points of contact and Let C be the centre.

At C, Reaction forces $R_B , R_A$ and weight W are under equilibrium

$\therefore \dfrac {R_A}{\sin (180-30)}=\dfrac {R_B}{\sin (180-50)} =\dfrac {500}{\sin (30+50)} \\ \therefore \dfrac {R_A}{\sin (150)} =\dfrac {R_B}{\sin (130)} =\dfrac {500}{\sin (80)} \\ \therefore R_A =\dfrac {500\sin 150}{\sin 80} =253.8567 N \\ \therefore R_B =\dfrac {500\sin 130}{\sin 80} =388.9310 N $

Reaction at $A=R_A=253.8567 N(50^\circ)$

Reaction at $B=R_B= 388.9310 N(30^\circ)$