Question

AB is the diameter of a circle, centre O, C is a point on the circumference such that ∠COB = $\theta$. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that

$\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\mathrm{\pi }\left(\frac{1}{2}-\frac{\mathrm{\theta }}{120}\right)$

Answer 1

We know that the area of minor segment of angle θ in a circle of radius r is,

.

It is given that,

So,

Area, A of minor segment cutoff by AC at angle

Now, since and

We know that the area of sector of a circle of radius r at an angle θ is

So, the area of sector BOC,

It is given that,