Question

Two parallel chords are drawn on the same side of the center of a circle of radius R. It is found that chords is

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

Answer 1

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

Let perpendicular distance from center to side opposite to angle $A$ be $x$

And let the perpendicular distance from center to side opposite to angle $2A$ be $x-y$

Applying trigonometry on triangle with $A$

$\cos \frac{A}{2}=\frac{x}{r}⟹x=r\cos \frac{A}{2}\cdots \left(1\right)$

On triangle with $2A$

$\cos A=\frac{x-y}{r}⟹x-y=r\cos A\cdots \left(2\right)$

Subtracting $\left(2\right)$ and $\left(1\right)$

$y=r(\cos \frac{A}{2}-\cos A)=2r\sin \frac{3A}{4}\sin \frac{A}{4}$

For $A=\theta ,2\theta$

Perpendicular distance between two parallel chords is $2r\sin \frac{3\theta }{2}\sin \frac{\theta }{2}$ and $2r\sin \frac{3\theta }{4}\sin \frac{\theta }{4}$