Let lines $A B$ and $C D$ intersect at $O .$ If $∠ A O C + ∠ B O E = 70^{\circ}$ and $∠ B O D = 40^{\circ},$ find $∠ B O E$ .

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Solution

Refer Image
$∠ A O G + ∠ B O E = 70^{\circ}$
$∠ D O B = ∠ A O C = 40^{\circ}$
$∴ 40^{\circ} + ∠ B O E = 70^{\circ}$
$∠ B O C = 70 - 40$
$= 30^{\circ}$

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Similar questions

In the figure below, lines AB and CD intersect at O. If $∠$ AOC + $∠$ BOE = $70^{\circ}$ and $∠$ BOD = $40^{\circ}$ , find (2 $∠$ BOE).

In the figure, lines $A B$ and $C D$ intersect at $O$ . If $∠ A O C + ∠ B O E = 70^{\circ}$ and $∠ B O D = 40^{\circ}$ , find $∠ B O E$ and reflex $∠ C O E$ .

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

Q. In the figure, lines

A B

and

C D

intersect at

O

. If

∠ A O C + ∠ B O E = 70

and

∠ B O D = 40

, find

∠ B O E

and reflex

∠ C O E

In the figure below, lines AB and CD intersect at O. If $∠ A O C$ + $∠ B O E$ = $70^{\circ}$ and $∠ B O D = 40^{\circ}$ , find $3 \times ∠ B O E$ in degrees.

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Let lines AB and CD intersect at O. If ∠AOC+∠BOE=70∘ and ∠BOD=40∘, find ∠BOE.

Refer Image∠AOG+∠BOE=70∘∠DOB=∠AOC=40∘∴40∘+∠BOE=70∘∠BOC=70−40=30∘

Let lines $A B$ and $C D$ intersect at $O .$ If $∠ A O C + ∠ B O E = 70^{\circ}$ and $∠ B O D = 40^{\circ},$ find $∠ B O E$ .

Refer Image
$∠ A O G + ∠ B O E = 70^{\circ}$
$∠ D O B = ∠ A O C = 40^{\circ}$
$∴ 40^{\circ} + ∠ B O E = 70^{\circ}$
$∠ B O C = 70 - 40$
$= 30^{\circ}$