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

Solution:

$\mathrm{AB}$ is a straight line, rays $\mathrm{OC}$ and $\mathrm{OE}$ stand on it.

$\therefore \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}$

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

$\Rightarrow 70^{\circ}+\angle \mathrm{COE}=180^{\circ}$

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

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

$\mathrm{CD}$ is a straight line, rays $\mathrm{OE}$ and $\mathrm{OB}$ stand on it.

$\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}+\angle \mathrm{BOD}=180^{\circ}$

$\Rightarrow 110^{\circ}+\angle \mathrm{BOE}+40^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{BOE}=180^{\circ}-150^{\circ}=30^{\circ}$