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Q.

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 $θ$ and 2 $θ$ at the centre of the circle. The perpendicular distance between the chords is

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a

$2 R \sin \frac{3 θ}{2} \sin \frac{θ}{2}$

b

$(1 - \cos \frac{θ}{2}) (1 + 2 \cos \frac{θ}{2}) R$

c

$(1 + \cos \frac{θ}{2}) (1 - 2 \cos \frac{θ}{2}) R$

d

$2 R \sin \frac{3 θ}{4} \sin \frac{θ}{4}$

answer is B, D.

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Detailed Solution

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$\begin{array}{l} OM = p_{1} = Rcos \frac{θ}{2} \\ ON = p_{2} = Rcos θ \\ MN = p_{1} - p_{2} = R (\cos \frac{θ}{2} - \cos θ) \\ = R 2 \sin \frac{3 θ}{4} \sin \frac{θ}{4} (d) \end{array}$
Again convert $\cos θ = 2 \cos^{2} \frac{θ}{2} - 1$ and factorise, we get
$= R (1 - \cos θ / 2)^{2} (1 + 2 \cos θ / 2)$

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