In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?

(c) 30°

In $\triangle O A B$, we have:

$\angle O A B+\angle O B A+\angle A O B=180^{\circ} \quad$ [Sum of the angles of a triangle]

$\Rightarrow 75^{\circ}+55^{\circ}+\angle A O B=180^{\circ}$

$\Rightarrow \angle A O B=50^{\circ}$

$\therefore \angle C O D=\angle A O B=50^{\circ} \quad$ [Vertically-Opposite Angles]

In $\Delta O C D$, we have:

$\angle C O D+\angle O C D+\angle O D C=180^{\circ} \quad[$ Sum of the angles of a triangle $]$

$\Rightarrow 50^{\circ}+100^{\circ}+\angle O D C=180^{\circ}$

$\Rightarrow \angle O D C=30^{\circ}$