Multiple choice

Let AB be a sector of a circle with centre O and radius d. $\angle$AOB $=\theta\left(< \displaystyle\frac{\pi}{2}\right)$, 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?

  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{\theta}{2}$
  3. $\displaystyle\frac{1}{2}\sin\theta$
  4. $\displaystyle{\sin^{-1}\left(\displaystyle\frac{1}{2}\sin\theta\right)}$
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D Correct answer
AI explanation

Place O at the origin so OA lies along the x-axis. In triangle OBD, the length OD is d cos theta and BD is d sin theta. Since E is the midpoint of BD, its y-coordinate is half of BD, which is (d sin theta)/2. The point F lies on the circle, so its y-coordinate is d sin(phi) where phi is the angle AOF; equating the y-coordinates because EF is parallel to OA gives d sin(phi) = (d sin theta)/2. Thus phi = sin^-1((sin theta)/2), and dividing this angle by theta gives the ratio of the arc AF to the arc AB as sin^-1((sin theta)/2) / theta. Wait, the required ratio of the length of arc AF to the length of arc AB is simply the measure of angle AOF, which is sin^-1((sin theta)/2). The answer is sin^-1((sin theta)/2).

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