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Let $\mathrm{AB}$ be a sector of a circle with centre $\mathrm{O}$ and radius d, $\angle \mathrm{AOB}=\theta\left( < \frac{\pi}{2}\right)$, and $\mathrm{D}$ be a point on $\mathrm{OA}$ such that $\mathrm{BD}$ is perpendicular $\mathrm{OA}$. Let $\mathrm{E}$ be the midpoint of $\mathrm{BD}$ and $\mathrm{F}$ be a point on the arc $\mathrm{AB}$ such that $\mathrm{EF}$ is parallel to $\mathrm{OA}$. Then the ratio of length of the arc AF to the length of the arc $\mathrm{AB}$ is -
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Verified Answer
The correct answer is:
$\frac{\sin ^{-1}\left(\frac{1}{2} \sin \theta\right)}{\theta}$
$\mathrm{CF}=\mathrm{r}(1-\cos \theta \sec \phi)$
$\mathrm{EC}=\mathrm{r}(\sin \phi-\cos \theta \tan \phi)$
$\mathrm{CD}=\mathrm{r} \cos \theta \tan \phi$
$\mathrm{EC}+\mathrm{CD}=\mathrm{ED}$
$\mathrm{r} \sin \phi=\frac{\mathrm{r} \sin \theta}{2}$
$\phi=\sin ^{-1}\left(\frac{\sin \theta}{2}\right)$
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