QJEEAdv10Ma6be5: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma6be5 Solution: Two parallel chords of a circle of radius 2 are at a distance…

Let $\theta = {\pi \over {2k}}$

$\cos \theta = {x \over 2}$

$\Rightarrow \cos 2\theta = {{\sqrt 3 + 1 - x} \over 2}$

$\Rightarrow 2{\cos ^2}\theta - 1 = {{\sqrt 3 + 1 - x} \over 2}$

$\Rightarrow 2\left( {{{{x^2}} \over 4}} \right) - 1 = {{\sqrt 3 + 1 - x} \over 2}$

$\Rightarrow {x^2} + x - 3 - \sqrt 3 = 0$

$\Rightarrow x = {{ - 1 \pm \sqrt {1 + 12 + 4\sqrt 3 } } \over 2}$

$= {{ - 1 \pm \sqrt {13 + 4\sqrt 3 } } \over 2}$

$= {{ - 1 + 2\sqrt 3 + 1} \over 2} = \sqrt 3$

$\therefore$ $\cos \theta = {{\sqrt 3 } \over 2} \Rightarrow \theta = {\pi \over 6}$

$\therefore$ Required angle $= {\pi \over k} = 2\theta = {\pi \over 3}$

$\Rightarrow k = 3$