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. ${\sin }^{-1}\left(\frac{1}{2}\sin \theta \right)$

Answer 1

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

$AB=lAF=m$

$\theta =\frac{l}{d},\alpha =\frac{m}{d}\Rightarrow \frac{m}{l}=\frac{\alpha }{\theta }$

Let P be intersection of $OF$ with $BD$. Let $\angle DOP$ be $\alpha$.

$\angle ODP=\angle PEF={90}^{\circ }$

$\angle DOP=\angle PFE=\alpha$

$BD=d\left(sin\theta \right)\Rightarrow DE=\frac{1}{2}dsin\theta$

Both $\Delta ODP$ are $\Delta PEF$ are right-angled triangles. From the diagram,

$OP\left(sin\alpha \right)+(d-OP)\left(sin\alpha \right)=DP+PE=DE=\frac{1}{2}dsin\theta$

$\Rightarrow \alpha =si{n}^{-1}\left(\frac{1}{2}sin\theta \right)$