area of shaded region in circle 

Solution : $\angle O = \angle D = \theta$ (corresponding angles)
Also we can use area of sector formula

After that, I have no idea
 A: The area of the shaded region $CDB$ is equal to the area of the sector $AOB$ minus the  sum of the areas of the sector $AOC$ and the triangle $\Delta OCD$:
$$A(CDB) = A(AOB) - A(AOC) - A(\Delta OCD)$$
The area of the sector $AOB$ is simply $(1/2) R^2 \theta$.
Now, to find the other areas, we need $\beta=\angle COD$.  We find this using the law of sines for the angle $\alpha = \angle OCD$:
$$\frac{\sin{(\pi-\theta)}}{R}=\frac{\sin{\alpha}}{\ell}$$
Combining this with $\alpha+\beta+(\pi-\theta)=\pi$ from $\Delta OCD$, we get that
$$\beta = \theta-\arcsin{\left (\frac{\ell }{R}\sin{\theta}\right )}$$
In terms of $\beta$, $A(AOC)=(1/2) R^2 (\theta-\beta)$ and $A(\Delta OCD)=(1/2) R \ell \sin{\beta}$.  Putting this all together, I get
$$A(CDB) = \frac12 R^2 \left [ \theta-\arcsin{\left (\frac{\ell }{R}\sin{\theta}\right )}\right] +\frac14 \ell^2 \sin{2 \theta} - \frac12 \ell \sin{\theta} \sqrt{R^2-\ell^2 \sin^2{\theta}}$$
A: Hint: The area of the shaded region equals the area of the sector $OCB$ minus the area of the triangle $OCD$.
