Question

In the given figure two lines $AB$ and $CD$ intersect at $O,$ $OE$ is any ray such that $\angle BOE={80}^{\circ }$. If $\angle BOD={30}^{\circ }$, then the find $\angle EOC$.

Answer 1

$\angle BOE+\angle AOE={180}^{\circ }$ (Supplementary Angles )

$\angle AOE={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

and $\angle BOD=\angle AOC={30}^{\circ }$ (Vertically Opposite Angles)

Hence, $\angle EOC={100}^{\circ }+{30}^{\circ }={130}^{\circ }$