In the given figure, AB || CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ?

(a) 130°

Draw $O E\|A B\| C D$

Now, $O E^{\prime} \mid A B$ and $O A$ is the transversal.

$\therefore \angle O A B+\angle A O E=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary]

$\Rightarrow \angle O A B+\angle A O C+\angle C O E=180^{\circ}$

$\Rightarrow 100^{\circ}+30^{\circ}+\angle C O E=180^{\circ}$

$\Rightarrow \angle C O E=50^{\circ}$

Also, 

$O E \| C D$ and $O C$ is the transversal.

$\therefore \angle O C D+\angle C O E=180^{\circ} \quad[$ Angles on the same side of a transversal line are supplementary $]$

$\Rightarrow \angle O C D+50^{\circ}=180^{\circ}$

$\Rightarrow \angle O C D=130^{\circ}$