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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 $k$ ]
[Note : $[k]$ denotes the largest integer less than or equal to $k$ ]
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1643 Upvotes
Verified Answer
The correct answer is:
3
Let $\theta=\frac{\pi}{2 k}$
$$
\begin{aligned}
& \cos \theta=\frac{x}{2} \\
& \Rightarrow \quad \cos 2 \theta=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2\left(\frac{x^2}{4}\right)-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow \quad x^2+x-3-\sqrt{3}=0 \\
& \Rightarrow \quad x=\frac{-1 \pm \sqrt{1+12+4 \sqrt{3}}}{2} \\
& =\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\
& =\frac{-1+2 \sqrt{3}+1}{2}=\sqrt{3} \\
& \therefore \quad \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \\
& \therefore \quad \text { Required angle }=\frac{\pi}{k}=2 \theta=\frac{\pi}{3} \\
& \Rightarrow \quad k=3 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \cos \theta=\frac{x}{2} \\
& \Rightarrow \quad \cos 2 \theta=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2\left(\frac{x^2}{4}\right)-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow \quad x^2+x-3-\sqrt{3}=0 \\
& \Rightarrow \quad x=\frac{-1 \pm \sqrt{1+12+4 \sqrt{3}}}{2} \\
& =\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\
& =\frac{-1+2 \sqrt{3}+1}{2}=\sqrt{3} \\
& \therefore \quad \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \\
& \therefore \quad \text { Required angle }=\frac{\pi}{k}=2 \theta=\frac{\pi}{3} \\
& \Rightarrow \quad k=3 \\
&
\end{aligned}
$$
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