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Q.
The acute angle between the lines joining the origin to the points of intersection of the line $ \sqrt{3}x+y=2 $ and the circle is $ {{x}^{2}}+{{y}^{2}}=4, $ is
Equation of line is $ \sqrt{3}x+y=2 $ ...(i) and equation of circle is $ {{x}^{2}}+{{y}^{2}}=4 $ ...(ii) From Eqs. (i) and (ii), we get $ {{x}^{2}}+2{{(-\sqrt{3}x)}^{2}}=4 $
$ \Rightarrow $ $ {{x}^{2}}+4+3{{x}^{2}}-4\sqrt{3}x=4 $
$ \Rightarrow $ $ 4{{x}^{2}}-4\sqrt{3}x=0 $
$ \Rightarrow $ $ x(x-\sqrt{3})=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} $ And slope of line joining (0, 0) and $ (\sqrt{3},-1) $
$=\frac{-1}{\sqrt{3}}\Rightarrow {{\theta }_{2}}=\frac{\pi }{6} $ .
Required angle
$=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3} $