Let OA be a radius of a circle with center O | vedclass

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(D) Given $\angle AOB = 	heta$ and radius $OF = OA = OB = d$.In $	riangle ODB$,$BD = OB \sin 	heta = d \sin 	heta$.Since $E$ is the mid-point of $

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$\frac{\sin^{-1}(\frac{1}{2} \sin 	heta)}{	heta}$

Vedclass · Answer

(D) Given $\angle AOB = 	heta$ and radius $OF = OA = OB = d$.In $	riangle ODB$,$BD = OB \sin 	heta = d \sin 	heta$.Since $E$ is the mid-point of $BD$,$ED = \frac{1}{2} BD = \frac{d}{2} \sin 	heta$.Let $F$ be a point on the arc $AB$ such that $EF \parallel OA$. Let $FM \perp OA$ where $M$ is on $OA$. Then $FM = ED = \frac{d}{2} \sin 	heta$.In $	riangle OFM$,$\sin \alpha = \frac{FM}{OF} = \frac{\frac{d}{2} \sin 	heta}{d} = \frac{1}{2} \sin 	heta$,where $\alpha = \angle AOF$.Thus,$\alpha = \sin^{-1}(\frac{1}{2} \sin 	heta)$.The length of arc $AF = d \alpha$ and the length of arc $AB = d 	heta$.The ratio of the length of arc $AF$ to the length of arc $AB$ is $\frac{d \alpha}{d 	heta} = \frac{\alpha}{	heta} = \frac{\sin^{-1}(\frac{1}{2} \sin 	heta)}{	heta}$.

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