Solve the following :

$\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}-\frac{1}{2} \mathrm{mu}^{2}=-\mathrm{mgh}$

$v^{2}=3 g l-2 g l(1+\cos \theta) \ldots \ldots \ldots 1$

and

$\left(m v^{2}\right) / l=m g \cos \theta$

$\mathrm{v}^{2}=\mathrm{gl} \cos \theta \ldots \ldots \ldots 2$

from equation 1 and 2

$3 g l-2 g l-2 g l \cos \theta=g l \cos \theta$

$\theta=\cos ^{-1}\left(\frac{1}{3}\right)$

Thus,

Angle $=180^{\circ}-\cos ^{-1}\left(\frac{\frac{1}{3}}{3}\right)$

Angle $=\cos ^{-1}(-1 / 3)$