In Fig. 6.13, 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} $.
"
Given:
Lines $AB$ and $CD$ intersect at $O$.
$\angle AOC + \angle BOE = 70^o$ and $\angle BOD = 40^o$
To do:
We have to find $\angle BOE$ and reflex $\angle COE$.
Solution:
$AOB$ is a line.
Therefore,
$\angle AOC + \angle COE + \angle BOE = 180^o$
$(\angle AOC + \angle BOE) + \angle COE = 180^o$
$70^o + \angle COE = 180^o$
$\angle COE = 180^o-70^o= 110^o$
$\angle AOC = \angle BOD = 40^o$ (Vertically opposite angles)
$\angle AOC + \angle BOE = 70^o$
This implies,
$\angle BOD + \angle BOE = 70^o$
$\angle BOE = 70^o - 40^o = 30^o$
Reflex $\angle COE = 360^o - \angle COE$
$= 360^o- 110^o$
$= 250^o$
Hence, $\angle BOE = 30^o$ and Reflex $\angle COE = 250^o$.
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