In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD = ?

(b) 115°

We have:

$\angle A O C=\angle B O D \quad[$ Vertically-Opposite Angles $]$

$\therefore \angle A O C+\angle B O D=130^{\circ}$

$\Rightarrow \angle A O C+\angle A O C=130^{\circ} \quad[\because \angle A O C=\angle B O D]$

$\Rightarrow 2 \angle A O C=130^{\circ}$

$\Rightarrow \angle A O C=65^{\circ}$

Now,

$\angle A O C+\angle A O D=180^{\circ}[\because C O D$ is a straight line $]$

$\Rightarrow 65^{\circ}+\angle A O D=180^{\circ}$

$\Rightarrow \angle A O C=115^{\circ}$