Find the reactions $R_A$ and $R_B$ at the points of contact.

Alpha (∝) = 20 degrees

Beta (β) = 60 degrees

The above diagram can be redrawn as a free Body Diagram as follows :-

Applying Lami’sTheorem :

$\dfrac W{\sin⁡(∝+β)}= \dfrac {R_A}{\sin⁡(180-β)}= \dfrac { R_B}{\sin⁡(180-α) } \\ ∴\dfrac W{\sin⁡(20+60)} = \dfrac { R_A}{\sin⁡(180-60) }=\dfrac { R_B}{\sin⁡(180-20) }\\ ∴\dfrac W{\sin⁡(80) } = \dfrac { R_A}{\sin⁡(120) } = \dfrac { R_B}{\sin⁡(160) } \\ ∴ R_A=(\dfrac {\sin120}{\sin 80)}W=0.879W \\ ∴ R_B=(\dfrac {\sin160}{\sin 80)}W=0.347W $