Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at…
Solution

$ \begin{aligned} & \cos \theta=\frac{x}{2} \\ & \Rightarrow \quad \cos 2 \theta=\frac{\sqrt{3}+1-x}{2} \\ & \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}+1-x}{2} \\ & \Rightarrow 2\left(\frac{x^2}{4}\right)-1=\frac{\sqrt{3}+1-x}{2} \\ & \Rightarrow \quad x^2+x-3-\sqrt{3}=0 \\ & \Rightarrow \quad x=\frac{-1 \pm \sqrt{1+12+4 \sqrt{3}}}{2} \\ & =\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\ & =\frac{-1+2 \sqrt{3}+1}{2}=\sqrt{3} \\ & \therefore \quad \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \\ & \therefore \quad \text { Required angle }=\frac{\pi}{k}=2 \theta=\frac{\pi}{3} \\ & \Rightarrow \quad k=3 \\ & \end{aligned} $
Asked in: JEE Advanced 2010 (Paper 2)