A pendulum is vibrating with an angular amplitude of $90°$ as shown in figure, then

At $\theta = 90° , v = 0, \therefore a_n=\frac{v^2}{r}=0$

and $at = g \sin 90° = g.$

At any angular position $\theta , h = l \cos \theta$

$$\begin{gathered} v^2 = 2gl \cos \theta \\ \to a_c=\frac{v^2}{r}=\frac{ 2gl \cos \theta }{l}= 2g \cos \theta \end{gathered}$$

and  $a_t= g \cos \theta$

$$\begin{gathered} \to \tan \theta = \frac{a_c}{a_t} \\ \to \frac{\sin \theta }{\cos \theta }=\frac{2g \cos \theta }{g\sin \theta } \\ \Rightarrow \cos \theta =\frac{1}{\sqrt{3}} \end{gathered}$$