Question

In the given figure, AB || CD. If ∠ABO = 130° and ∠OCD = 110, then ∠BOC = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer 1

(b) 60°

Draw $EOF\parallel AB\parallel CD$ through O.
Now, $EO\parallel AB\text{and}OB$ is the transversal.
$$\begin{gathered} \therefore \angle ABO+\angle EOB=180°\left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow 130°+\angle EOB=180° \\ \Rightarrow \angle EOB=50° \end{gathered}$$

Also, $OF\parallel CD\text{and}OC$ is the transversal.
$$\begin{gathered} \therefore \angle OCD+\angle COF=180°\left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow 110°+\angle COF=180° \\ \Rightarrow \angle COF=70° \\ \text{Now,} \\ \angle EOB+\angle BOC+\angle COF=180°\left[\because EOF\mathit{}\text{is a straight line}\right] \\ \Rightarrow 50°+\angle BOC+70°=180° \\ \Rightarrow \angle BOC=\mathbf{60}\mathbf{°} \end{gathered}$$