# 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 $\angle \mathrm{COE}$."

#### Complete Python Prime Pack

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack

9 Courses     2 eBooks

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 $\angle COE = 250^o$.

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