Multiple choice

Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of $\theta $ and $2\theta $ at the centre of the circle. The perpendicular distance between the chords is

  1. $1R\sin \dfrac{{3\theta }}{2}\sin \dfrac{\theta }{2}$
  2. $\left( {1 + \cos \dfrac{\theta }{2}} \right)\left( {1 - 2\cot \dfrac{\theta }{1}} \right)R$
  3. $2R\sin \dfrac{{3\theta }}{4}\sin \dfrac{\theta }{4}$
  4. $2R\sin 3\theta $
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C Correct answer
Explanation

The distance from the center to a chord subtending angle 2A is R*cos(A). For angles theta and 2*theta, the distances are R*cos(theta/2) and R*cos(theta). The difference is R(cos(theta/2) - cos(theta)). Using trigonometric identities, this simplifies to 2R*sin(3*theta/4)*sin(theta/4).

AI explanation

The perpendicular distances from the center to the two chords are $R\cos(\theta/2)$ and $R\cos(2\theta/2)$, which is $R\cos\theta$. The distance between the chords is the difference of these distances, giving $R\cos(\theta/2) - R\cos\theta$. By the sum-to-product identity, $\cos A - \cos B = -2\sin((A+B)/2)\sin((A-B)/2)$, so we get $-2R\sin(3\theta/4)\sin(-\theta/4)$. This simplifies to $2R\sin(3\theta/4)\sin(\theta/4)$.

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