Answer
$\dfrac{\pi}{6}$
Work Step by Step
Suppose $\theta = \cos^{-1} (\dfrac{\sqrt 3}{2})$
This gives: $\cos \theta= \dfrac{\sqrt 3}{2}$
Now, $\theta = \cos^{-1} (\dfrac{\sqrt 3}{2})=\dfrac{\pi}{6}$
Therefore, $\cos^{-1} (\dfrac{\sqrt 3}{2})=\dfrac{\pi}{6}$
