Question

In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD = ?
(a) 65°
(b) 115°
(c) 110°
(d) 125°

Answer 1

(b) 115°

We have :
$$\begin{gathered} \angle AOC=\angle BOD\left[\text{Vertically-Opposite Angles}\right] \\ \therefore \angle AOC+\angle BOD=130° \\ \Rightarrow \angle AOC+\angle AOC=130°\left[\because \angle AOC=\angle BOD\right] \\ \Rightarrow 2\angle AOC=130° \\ \Rightarrow \angle AOC=65° \end{gathered}$$
Now,
$$\begin{gathered} \angle AOC+\angle AOD=180°\left[\text{∵}\mathit{\text{COD}}\text{is a straight line}\right] \\ \Rightarrow 65°+\angle AOD=180° \\ \Rightarrow \angle AOC=115° \end{gathered}$$