Question

In the given figure, $\angle OAB={110}^{\circ }$ and $\angle BCD={130}^{\circ }$ then $\angle ABC$ is equal to

(a) ${40}^{\circ }$

(b) ${50}^{\circ }$

(c) ${60}^{\circ }$

(d) ${70}^{\circ }$

Answer 1

In the given figure, $OA||CD$.
Construction: Extend $OA$ such that it intersects $BC$ at $E$.
Now, $OE||CD$ and $BC$ is a transversal.
$\therefore \angle AEC=\angle BCD=130°$ (Pair of corresponding angles)
Also,$\angle OAB+\angle BAE=180°$ (Linear pair)
$\therefore 110°+\angle BAE=180°$
$\Rightarrow \angle BAE=180°-110°=70°$
In $\Delta ABE$,
$\angle AEC=\angle BAE+\angle ABE$
[In a triangle, exterior angle is equal to the sum of two opposite interior angles]
$\therefore 130°=70°+x°$
$\Rightarrow x°=130°-70°=60°$
Thus, the measure of angle $\angle ABC$ is $60°.$
Hence, the correct answer is option (c).