Question

In the given figure, AOB is a straight line. If $\angle AOC=(3x-10{)}^{\circ }$, \angle COD = 50^{\circ}\) and $\angle BOD=(x+20{)}^{\circ }$ then $\angle AOC=?$

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

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

(c) ${80}^{\circ }$

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

Answer 1

ANSWER:
(c) 80°
We have :
∠AOC+∠COD+∠BOD=180° Since AOB is a straight line ⇒3x-10+50+x+20=180⇒4x=120⇒x=30