Explanation

(a) Let $A$ be the angle of prism, and $\delta\ $the angle of minimum deviation, then refractive index of the medium of prism is given by
$\mu = \frac{\sin\left( \frac{A + \delta}{2} \right)}{\sin\left( \frac{A}{2}\ \right)}$
Given, $\delta = A,\ \mu = 1.5$
$\therefore 1.5 = \frac{\sin\left( \frac{A + A}{2} \right)}{\sin\left( \frac{A}{2} \right)}$
Also sin 2$\theta = 2$ sin$\theta$ cos$\theta$
$\therefore 1.5 = \frac{2sin\frac{A}{2}\cos\frac{A}{2}}{sin\frac{A}{2}}$
$\Longrightarrow \cos{\frac{A}{2} = \frac{1.5}{2} = 0.75}\ $
$\Longrightarrow \frac{A}{2} = 41{^\circ} - 24^{'} - 36^{''}$
$\Longrightarrow A = 82{^\circ} - 48^{'} - 72''$