Question

In Fig. 8.62, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40° find ∠BOE and reflex ∠COE.

Answer 1

∠BOE = 30° and reflex∠COE = 110°

Explanation:

Given data:

AB and CD are intersecting lines.

∠AOC + ∠BOE = 70° and ∠BOD = 40°

Sum of angle in linear pair = 180°

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

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

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

⇒ ∠COE = 110°

Sum of angle in linear pair = 180°

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

$110^{\circ}+\angle B O E+40^{\circ}=180^{\circ}$

$150^{\circ}+\angle B O E=180^{\circ}$

$\angle B O E=180^{\circ}-150^{\circ}$

⇒ ∠BOE = 30°

To learn more...

In fig 6.13 lines AB and CD intersect at O.if Angle AOC+AngleBOE =70° and Angle BOD =40° ,find ANGLE BOE and Reflex Angle COE.

https://brainly.in/question/3121202

Answer 2

$\huge{Question\::-}$

In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE

$\huge{Solution\::-}$

Since AB is a straight line,

• ∠AOC + ∠COE + ∠EOB = 180°

• or (∠AOC + ∠BOE) + ∠COE = 180°

• or 70° + ∠COE = 180°

• [ ∵∠AOC + ∠BOE = 70° (Given)]

• or ∠COE = 180° – 70° = 110°

⠀⠀∴ Reflex ∠COE

⠀⠀ = 360° – 110°

⠀⠀ = 250°

Also, AB and CD intersect at O.

⠀⠀⠀∴∠COA = ∠BOD

⠀⠀⠀⠀⠀⠀⠀ [Vertically opposite angles]

⠀⠀⠀But ∠BOD = 40°

⠀⠀⠀⠀⠀⠀⠀ [Given]

⠀⠀⠀∴ ∠COA = 40°

⠀⠀⠀Also, ∠AOC + ∠BOE

⠀⠀⠀= 70°

________

• ∴ 40° + ∠BOE = 70°

• or ∠BOE = 70° -40°

⠀⠀ = 30°

Thus, ∠BOE = 30°and,

reflex ∠COE = 250°.