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Two lines drawn through a point on the circumference of a circle divided the circle into three regions of equal area.Then the angle $\theta$ between the line is given by

$\begin{array}{1 1}(a)\;3\theta+3\sin\theta=\pi&(b)\;6\theta+3\sin\theta=\pi\\(c)\;2\theta+\sin\theta=\pi&(d)\;\theta+\sin\theta=\large\frac{\pi}{2}\end{array}$

Area of region $ABC=\large\frac{\pi r^2}{3}$
Area of $OAB=\large\frac{1}{2}$$r^2.2\theta \qquad\qquad\quad=r^2\theta Area of \Delta OAC=\large\frac{1}{2}$$r^2\sin\theta$=Area of $\Delta OBC$
$\Rightarrow \large\frac{1}{2}$$r^2\sin\theta+\large\frac{1}{2}$$r^2\sin\theta+r^2\theta$
$\Rightarrow \large\frac{\pi r^2}{3}$
$\Rightarrow 3\sin\theta+3\theta=\pi$
Hence (a) is the correct answer.
edited Mar 3, 2014
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