Assume mass of each part = m

Here, $${{{C_1}R} \over {{C_1}A}} = \sin \theta $$

$$ \therefore $$   C₁R = C₁A(sin$$\theta $$)

= $${1 \over 2}$$ sin$$\theta $$

$${{BY} \over {AB}} = \sin \theta $$

$$ \Rightarrow $$   BY = Lsin$$\theta $$

As By = MP

$$ \therefore $$   MP = Lsin$$\theta $$

$${{{C_2}M} \over {B{C_2}}}$$ = cos$$\theta $$

$$ \Rightarrow $$   C₂M = BC₂cos$$\theta $$

$$ \Rightarrow $$   C₂M = $${L \over 2}\cos \theta $$

$$ \therefore $$   C₂P = C₂M $$-$$ MP = $${L \over 2}\cos \theta $$ $$-$$ Lsin$$\theta $$

Now balancing torque about hinge paint A,

mg(C₁R) = mg(C₂P)

$$ \Rightarrow $$   mg$$\left( {{L \over 2}\sin \theta } \right)$$ = mg$$\left( {{L \over 2}\cos \theta - L\sin \theta } \right)$$

$$ \Rightarrow $$   $${{\sin \theta } \over 2}$$ = $${{\cos \theta } \over 2} - $$ sin$$\theta $$

$$ \Rightarrow $$   $${{3\sin \theta } \over 2}$$ = $${{\cos \theta } \over 2}$$

$$ \Rightarrow $$   tan$$\theta $$ = $${1 \over 3}$$