Answer
a) $F_1=mgsin20^{\circ}$
b) $F_2=mgcos20^{\circ}$
Work Step by Step
(a) We can find the contact force 1 as
$\Sigma F_x=F_1-mgsin\theta=0$
$\implies F_1=mgsin\theta$
$\implies F_1=mgsin20^{\circ}$
(b) The contact force 2 is given as follows:
$\Sigma F_y=F_2-mgcos\theta=0$
$\implies F_2=mgcos\theta$
$\implies F_2=mgcos20^{\circ}$
