Two parallel chords of a circle of radius 2 a | vedclass

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(C) Let the radius of the circle be $r = 2$. The distance of a chord subtending an angle $	heta$ at the center from the center is $d = r \cos(	heta/

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$5$

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(C) Let the radius of the circle be $r = 2$. The distance of a chord subtending an angle $	heta$ at the center from the center is $d = r \cos(	heta/2)$.For the two chords,the angles subtended are $	heta_1 = \frac{\pi}{k}$ and $	heta_2 = \frac{2\pi}{k}$.The distances of these chords from the center are $d_1 = 2 \cos(\frac{\pi}{2k})$ and $d_2 = 2 \cos(\frac{\pi}{k})$.Since the chords are on the same side of the center (implied by the distance sum),the distance between them is $d_1 - d_2 = \sqrt{3} + 1$ (or $d_1 + d_2$ if on opposite sides).Given $2 \cos(\frac{\pi}{2k}) - 2 \cos(\frac{\pi}{k}) = \sqrt{3} + 1$ leads to no real $k > 0$ for the given range.Assuming the chords are on opposite sides of the center: $2 \cos(\frac{\pi}{2k}) + 2 \cos(\frac{\pi}{k}) = \sqrt{3} + 1$.Let $	heta = \frac{\pi}{k}$. Then $2 \cos(\frac{	heta}{2}) + 2 \cos(	heta) = \sqrt{3} + 1$.Using $\cos(	heta) = 2 \cos^2(\frac{	heta}{2}) - 1$,we get $2 \cos(\frac{	heta}{2}) + 2(2 \cos^2(\frac{	heta}{2}) - 1) = \sqrt{3} + 1$.$4 \cos^2(\frac{	heta}{2}) + 2 \cos(\frac{	heta}{2}) - 3 - \sqrt{3} = 0$.Solving for $\cos(\frac{	heta}{2})$ using the quadratic formula: $\cos(\frac{	heta}{2}) = \frac{-2 \pm \sqrt{4 - 16(-3 - \sqrt{3})}}{8} = \frac{-2 \pm \sqrt{52 + 16\sqrt{3}}}{8} = \frac{-2 \pm (2\sqrt{3} + 4)}{8}$.Taking the positive root: $\cos(\frac{	heta}{2}) = \frac{2\sqrt{3} + 2}{8} = \frac{\sqrt{3} + 1}{4}$. This does not yield a standard angle.Re-evaluating the distance: If $d_1 = 2 \cos(\frac{\pi}{2k})$ and $d_2 = 2 \cos(\frac{\pi}{k})$,and $d_1 + d_2 = \sqrt{3} + 1$,the solution $k=3$ is obtained if the equation was $2 \cos(\frac{\pi}{2k}) + 2 \cos(\frac{\pi}{k}) = \sqrt{3} + 1$ with specific values. For $k=3$,$\cos(\frac{\pi}{6}) + \cos(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2}$. Multiplying by $2$ gives $\sqrt{3}+1$. Thus $k=3$.$[k] = [3] = 3$.

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 $k$ ].

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