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

$Reaction\ forces\ R_A, R_B\ and\ weight\ 'W' are\ under\ equilibrium\, at\ centre\ point\ C.$

$Hence, \dfrac {R_A} {sin(180-30)} =\dfrac {R_B} {sin(180-50)} = \dfrac {500} {sin(30+50)} $

$Hence, \dfrac {R_A} {sin(150)} =\dfrac {R_B} {sin(130)} = \dfrac {500} {sin(80)} $

$ {R_A} =\dfrac {500sin(150)} {sin(80)}$

$= 253.8567\ N $

$ {R_B} =\dfrac {500sin(130)} {sin(80)}$

$= 388.9310\ N$

$Hence,\ the\ reaction\ at\ contact\ point\ A\ = R_A=253.8567\ N\ with\ \angle 50^\circ \\and\ the\ reaction\ at\ contact\ point\ B\ =R_B=388.9310\ N\ with\ \angle 30^\circ $