Question

In the given figure, AB || CD. If ∠ABO = 45° and ∠COD = 100°, ∠CDO = ?
(a) 25°
(b) 30°
(c) 35°
(d) 45°

Answer 1

(c) 35°

$AB\parallel CD\text{and}BOC$ is the transversal.
$\therefore \angle OCD=\angle OBA=45°\left[\text{Alternate Interior Angles}\right]$
Now, in $∆OCD$, we have:
$$\begin{gathered} \angle COD+\angle OCD+\angle CDO=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 100°+45°+\angle CDO=180° \\ \Rightarrow \angle CDO=\mathbf{35}\mathbf{°} \end{gathered}$$