Question

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

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

Answer: (A), (D)

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

Let perpendicular distance from center to side opposite to angle A be x.
and let perpendicular distance from center to side opposite to angle $2a$ be $x-y.$
Applying trigonometry on triangle with $A$.
$\cos \frac{A}{2}=\frac{x}{r}x=r\left(\cos \frac{A}{2}\right)⟶\left(1\right)$
On triangle with angle $2A.bv$
$\cos A=\frac{x-y}{2}x-y=r\left(\cos A\right)⟶\left(2\right)$
subtracting $\left(2\right)\&\left(1\right)y=r(\cos \frac{A}{2}-\cos A)$
$\therefore$ Perpendicular distance b/w two parallel chords is $r(\cos \frac{A}{2}-\cos A).$
On using further trigonometric functions, we get
$2r\sin \frac{3\theta }{2}\sin \frac{\theta }{2}2r\sin \frac{3\theta }{4}\sin \frac{\theta }{4}$