Question

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

• A. $1R\sin \frac{3\theta }{2}\sin \frac{\theta }{2}$
• B. $\left(1+\cos \frac{\theta }{2}\right)\left(1-2\cot \frac{\theta }{1}\right)R$
• C. $2R\sin \frac{3\theta }{4}\sin \frac{\theta }{4}$
• D. $2R\sin 3\theta$

Answer 1

The correct option is C $2R\sin \frac{3\theta }{4}\sin \frac{\theta }{4}$

we need to find $PQ$

Given:

$\angle AOB=2\theta$

$\angle COD=\theta$

$1Q=(OQ-OP)--\left(1\right)$

in $△POB\Rightarrow \angle POB=\frac{\angle AOB}{2}=\frac{2\theta }{2}=\theta$

$\cos \angle POB=\frac{OP}{OB}$

$\Rightarrow \cos \theta =\frac{OP}{OB}\Rightarrow OP=OB\cos \theta$

in $△QOD$

$\angle QOD=\frac{\angle COD}{2}=\frac{\theta }{2}$

$\cos \angle QOD=\frac{OQ}{OD}$

$\Rightarrow \cos \left(\theta /2\right)=\frac{OQ}{OD}$

$\Rightarrow OQ=OD\cos \left(\theta /2\right)$

$OB=OD=OC=OA=R$ (rod of circle)

from $\left(1\right),\left(2\right)$ & $\left(3\right)$

$PQ=r\cos \left(\theta /2\right)-r\cos \theta R$

$PQ=r[\cos \theta /2-\cos \theta ]$

$PQ=2r\sin \left(\frac{\theta /2+\theta }{2}\right)\sin \left(\frac{\theta -\theta /2}{2}\right)$

$PQ=2R\sin \left(\frac{3\theta }{4}\right)\sin \left(\theta /4\right)$