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

Answer 1

The area of the minor segment is given by $\pi r^2\times \frac{\theta }{360}-r^{2}sin\frac{\theta }{2}cos\frac{\theta }{2}=\frac{\pi r^2}{8}$

$\Rightarrow \frac{\pi \theta }{360}-sin\frac{\theta }{2}cos\frac{\theta }{2}=\frac{\pi }{8}$

$\Rightarrow \frac{\pi \theta }{45}-8sin\frac{\theta }{2}cos\frac{\theta }{2}=\pi$

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

Hence, proved.