Question

In the given figure, AB || CD.If $\angle BAO={60}^{\circ }$ and $\angle OCD={110}^{\circ }$ then $\angle AOC=?$

$\left(a\right){70}^{\circ }$

$\left(b\right){60}^{\circ }$

$\left(c\right){50}^{\circ }$

$\left(d\right){40}^{\circ }$

Answer 1

Now, EC ∥ AB and EA is the transversal.
∴ ∠OEC=∠OAB=60° (Corresponding angles)

Now, ∠OCE +∠OCD=180°
⇒∠OCE+110°=180°
⇒∠OCE=70°

In triangle OEC, ∠OEC+∠OCE+∠EOC=180°
⇒60°+70°+∠EOC=180°
⇒∠EOC=50°
⇒∠AOC=50°

So, the correct choice is (c).