Question

A chord of a circle subtends an angle of $\theta$ at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that

$8\sin \frac{\theta }{2}\cos \frac{\theta }{2}+\mathrm{\pi }=\frac{\mathrm{\pi \theta }}{45}.$

Answer 1

We know that the area of circle and area of minor segment of angle θ in a circle of radius r is given by, andrespectively.

It is given that,