Question

In the figure below, straight lines AB and CD intersect at O. If $\angle$AOC + $\angle$BOE = ${70}^{\circ }$ and $\angle$BOD = ${40}^{\circ }$, find (2$\angle$BOE).

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\angle$AOC + $\angle$BOE + $\angle$COE = ${180}^{\circ }$ (sum of angles on a straight line)

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 = ${180}^{\circ }$ (sum of angles on a straight line)

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

So, $\angle$BOE = ${180}^{\circ }$ – $\angle$COE - $\angle$BOD => ${180}^{\circ }$ – ${110}^{\circ }$ - ${40}^{\circ }$ = ${30}^{\circ }$
(2 $\angle$BOE) = ${60}^{\circ }$