Question

$AB$ is the diameter of a circle, centre $O$. $C$ is a point on the circumference such that $\angle COB=\theta$. The area of the minor segment cut off by $AC$ is equal to twice, the area of the sector $BOC$. Find whether the statement $\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\pi (\frac{1}{2}-\frac{\theta }{120})$ is True/False

• A. True
• B. False

Answer 1

The correct option is A True

From given figure, we have,

area of sector $BOC=\frac{\theta }{360}\times \pi r^2$

area of segment cut of by AC = area of sector - area of triangle AOC

area of sector $=\frac{180-\theta }{360}\times \pi r^2=\frac{\pi r^2}{2}-\frac{(\pi \theta r{)}^2}{360}$

area of segment $=\frac{\pi r^2}{2}-\frac{\pi \theta r^2}{360}-\frac{1}{2}(AC\times OM)$

$=\frac{\pi r^2}{2}-\frac{\pi \theta r^2}{360}-\frac{1}{2}\times \left(2R\cos \frac{\theta }{2}R\sin \frac{\theta }{2}\right)$

$=r^2[\frac{\pi }{2}-\frac{\pi \theta }{360}-\cos \frac{\theta }{2}\sin \frac{\theta }{2}]$

area of segment AC = 2(area of sector BDC)

$r^2[\frac{\pi }{2}-\frac{\pi \theta }{360}-\cos \frac{\theta }{2}\sin \frac{\theta }{2}]=2r^2\left[\frac{\pi \theta }{360}\right]$

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

$=\frac{\pi }{2}-\frac{\pi \theta }{360}[1+2]$

$=\pi (\frac{1}{2}-\frac{\theta }{120})$

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