(b)

Let $v$ is the velocity of bob at position $\theta$ after being released from horizontal position.

From energy conservation, we have

$m g h=\frac{1}{2} m v^2$

$\Rightarrow m g(l \cos \theta)=\frac{1}{2} m v^2$

$\Rightarrow \quad \frac{v^2}{l}=2 g \cos \theta \quad \dots(i)$

Restoring force on bob is component of weight.

So, tangential component of acceleration is

$a_t=g \sin \theta \quad \dots(ii)$

If total acceleration a makes angle $\phi$ with string. Then,

Tangential component,

$a_t=a \sin \phi$

and radial component, $a_c=a \cos \phi$

So, $\quad \frac{a_t}{a_c}=\frac{a \sin \phi}{a \cos \phi}=\tan \phi$

$\Rightarrow \quad \tan \phi=\frac{a_i}{a_c} \quad \dots(iii)$

Substituting values of $a_t$ and $a_c$ from Eqs.

$(i)$ and $(ii)$ in Eq. $(iii)$, we get

$\tan \phi=\frac{g \sin \theta}{2 g \cos \theta}$

$\Rightarrow \quad \tan \phi=\frac{\tan \theta}{2}$

So, $\quad \phi=\tan ^{-1}\left(\frac{\tan \theta}{2}\right)$