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$\angle$BOE).

• A. 30${}^{\circ }$
• B. 60${}^{\circ }$
• C. 90${}^{\circ }$
• D. 180${}^{\circ }$

Answer 1

The correct option is C

90${}^{\circ }$

$\angle$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${}^{\circ }$