Question

In the given figure, lines $AB$ and $CD$ intersect at $O$.If $\angle AOC+\angle BOE={70}^{\circ }$ and $\angle BOD={40}^{\circ }$ find $\angle BOE$ and reflex of $\angle COE$

Answer 1

Given:$\angle BOD={40}^{\circ }$

Since $AB$ and $CD$ intersects, $\angle AOC=\angle BOD$(vertically opposite angles)

$\angle AOC={40}^{\circ }$

Also,$\angle AOC+\angle BOE={70}^{\circ }$

$\Rightarrow \angle BOE={70}^{\circ }-\angle AOC={70}^{\circ }-{40}^{\circ }={30}^{\circ }$

We need to find reflex$\angle COE$

Reflex$\angle COE={360}^{\circ }-\angle COE$

Now, $\angle AOC+\angle COE+\angle BOE={180}^{\circ }$

$\Rightarrow \angle COE+(\angle AOC+\angle BOE)={180}^{\circ }$

$\Rightarrow \angle COE+({40}^{\circ }+{30}^{\circ })={180}^{\circ }$

$\Rightarrow \angle COE={180}^{\circ }-{70}^{\circ }={110}^{\circ }$

Reflex$\angle COE={360}^{\circ }-{110}^{\circ }={250}^{\circ }$