Question

In the given figure, AB || CD. If $\angle ABO={130}^0$ and $\angle OCD={110}^o$, then ∠BOC = ____.

• A. ${80}^{\circ }$
• B. ${100}^{\circ }$
• C. ${120}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$



Through O, draw EOF || AB || CD.
Now, AB || EO and BO is the transversal.
$\angle ABO+\angle EOB={180}^o$
$\Rightarrow {130}^o+\angle EOB={180}^o$
$\Rightarrow \angle EOB={50}^o$

Also,
CD || OF and CO is the transversal.
$\angle OCD+\angle COF={180}^o$
$\Rightarrow {110}^o+\angle COF={180}^o$
$\Rightarrow \angle COF={70}^o$

Let ∠BOC = x
Then,
$\angle EOB+\angle BOC+\angle COF={180}^o$ [Sum of angles on a straight line]
$\Rightarrow {50}^o+x+{70}^o={180}^o$
$\Rightarrow x=({180}^o-{120}^o)={60}^o$
So, $\angle BOC={60}^o$