Question

In the figure below, lines AB and CD intersect at O. If $\angle AOC$ + $\angle BOE$ = ${70}^{\circ }$ and $\angle BOD={40}^{\circ }$, find $3\times \angle BOE$ in degrees.

• A. 90

Answer 1

The correct option is A 90
$\angle AOC+\angle BOE+\angle COE={180}^{\circ }$ [Angles on a straight line equals ${180}^{\circ }$]

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

So, $\angle COE={180}^{\circ }-(\angle AOC+\angle COE)={180}^{\circ }-{70}^{\circ }={110}^{\circ }$

Again,
$\angle COE+\angle BOE+\angle BOD={180}^{\circ }$
[Angles on a straight line equals ${180}^{\circ }$]

We know that,
$\angle COE={110}^{\circ }$ and $\angle BOD={40}^{\circ }$

So, $\angle BOE={180}^{\circ }-\angle COE-\angle BOD={180}^{\circ }-{110}^{\circ }-{40}^{\circ }={30}^{\circ }$

$⟹$$3\times \angle BOE={90}^{\circ }$