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} $.
"

AcademicMathematicsNCERTClass 9

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 $\angle COE = 250^o$.

raja
Updated on 10-Oct-2022 13:40:27

Advertisements