Answer
$\theta_2 = 60^{\circ}$
Work Step by Step
Let $L$ be the length of the rod.
We can consider the torque about a rotation axis at the hinge:
$\sum \tau = 0$
$L~F_T~sin~\theta_2-(\frac{L}{2})~sin~60^{\circ}~mg= 0$
$L~(\frac{mg}{2})~sin~\theta_2 = (\frac{L}{2})~sin~60^{\circ}~mg$
$sin~\theta_2 = sin~60^{\circ}$
$\theta_2 = 60^{\circ}$
