$ \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta)$

$ \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta)$

Acceleration at lowest point $=\frac{\mathrm{v}^2}{\ell}$

Acceleration at extreme point $=g \sin \theta$

Hence, $\frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta$

$ \therefore \sin \theta=2(1-\cos \theta) $

$ \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$