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

Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of $\frac{\pi}{k}$ and $\frac{2 \pi}{k}$, where $k>0$, then the value of $[k]$ is [Note : $[k]$ denotes the largest integer less than or equal to $k$ ]

Solution

Let $\theta=\frac{\pi}{2 k}$
$ \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)

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