A chord of a circle subtends an angle of θ 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
$8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+\pi=\frac{\pi \theta}{45}$
We know that the area of circle and area of minor segment of angle $\theta$ in a circle of radius $r$ is given by, $A^{\prime}=\pi r^{2}$ and $A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$ respectively.
It is given that,
Area of minor segment $=\frac{1}{8} \times$ area of circle
$\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}=\frac{\pi r^{2}}{8}$
$\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} \times 8=\pi$
$\frac{8 \pi \theta}{360^{\circ}}-8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi$
$\frac{\pi \theta}{45^{\circ}}-8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi$
$\frac{\pi \theta}{45^{\circ}}=8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+\pi$
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