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 (2$\angle$BOE).

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\angle$AOC + $\angle$BOE + $\angle$COE = ${180}^{\circ }\left(Linearpairofangles\right).Given,\angle$AOC + $\angle$BOE = ${70}^{\circ }$.

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

Again, $\angle$COE + $\angle$BOE + $\angle$BOD = ${180}^{\circ }$ (Linear pair of angles).

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

So, $\angle$BOE = ${180}^{\circ }$ – $\angle$COE - $\angle$BOD $\Rightarrow$ ${180}^{\circ }$ – ${110}^{\circ }$ - ${40}^{\circ }$ = ${30}^{\circ }$.
Hence, 2 $\angle$BOE = ${60}^{\circ }$.