Explanation

(b)
Given equations of line and circle are respectively
$\sqrt{3}x + y = 2$ …(i)
$and\ x^{2} + y^{2} = 4$ …(ii)
From Eqs. (i) and (ii), we get
$x^{2} + (2 - \sqrt{3}x)^{2} = 4$
$\Rightarrow \ {4x}^{2} - 4\sqrt{3}x = 0$
$\Rightarrow \ x\left( x - \sqrt{3} \right) = 0\ \Rightarrow \ x = 0,\ \sqrt{3}$
$\therefore\$Points of intersection of line and circle are (0, 2) and $(\sqrt{3}, - 1)$.
Slope, of line joining (0, 0) and (0, 2)
$= \frac{2 - 0}{0 - 0} = \infty\ \Rightarrow \ \theta_{1} = \frac{\pi}{2}$
Also, slope of line joining (0, 0) and $\left( \sqrt{3}, - 1 \right)$
$= \frac{- 1}{\sqrt{3}}\ \Rightarrow \ \theta_{2} = \frac{\pi}{6}$
$\therefore Required\ angle = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$