Question

Let $AB$ be a sector of a circle with centre $O$ and radius $d$, $\angle AOB=\theta (<\frac{\pi }{2})$ , and $D$ be a point on $OA$ such that $BD$ is perpendicular $OA$. Let $E$ be the midpoint of $BD$ and $F$ be a point on the arc $AB$ such that $EF$ is parallel to $OA$. Then the ratio of length of the arc $AF$ to the length of the arc $AB$ is

• A. $\frac{1}{2}$
• B. $\frac{\theta }{2}$
• C. $\frac{1}{2}\sin \theta$
• D. $\frac{{\sin }^{-1}\left(\frac{1}{2}\sin \theta \right)}{\theta }$

Answer 1

The correct option is D $\frac{{\sin }^{-1}\left(\frac{1}{2}\sin \theta \right)}{\theta }$


From the figure,
$CD=OC\sin \varphi CE=CF\sin \varphi \Rightarrow CD+CE=(OC+CF)\sin \varphi \Rightarrow \frac{1}{2}BD=d\sin \varphi \Rightarrow \frac{d\sin \theta }{2}=d\sin \varphi \Rightarrow \sin \theta =2\sin \varphi \therefore \varphi ={\sin }^{-1}\left(\frac{1}{2}\sin \theta \right)$
Ratio of the arc length $\frac{AF}{AB}=\frac{\frac{\varphi }{2\pi }\times 2\pi r}{\frac{\theta }{2\pi }\times 2\pi r}=\frac{\varphi }{\theta }\Rightarrow \frac{AF}{AB}=\frac{{\sin }^{-1}\left(\frac{1}{2}\sin \theta \right)}{\theta }$