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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Mathematics
Chapter
Circle

JEE Advanced 2010 · Paper 2 · Mathematics · Circle

Two parallel chords of a circle of radius 2 are at a distance 3 +1...

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

Correct answer

0

Step-by-step solution

Let θ=π2k\theta=\frac{\pi}{2 k} Two parallel chords of a circle of radius 2 are at a distance 3 +1 apart. If the chords subtend at the centre, angles
cos⁡θ=x2⇒cos⁡2θ=3+1−x2⇒2cos⁡2θ−1=3+1−x2⇒2(x24)−1=3+1−x2⇒x2+x−3−3=0⇒x=−1±1+12+432=−1±13+432=−1+23+12=3∴cos⁡θ=32⇒θ=π6∴ Required angle =πk=2θ=π3⇒k=3\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 x²+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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