Question

In fig. AB and CD intersect each other at O. If <AOC + <BOE = 70° and <BOD = 40° then the value of <BOE is
1. 150°
2. 30°
3. 120°
4. 90°​

Answer 1

Answer:

∠BOE=30°

Step-by-step explanation:

From given figure

∠AOC=∠BOD [Vertically opposite angles]

⇒∠AOC=40 °

[∵∠BOD=40 °is given]

Now, ∠AOC+∠BOE=70 ° [Given]

⇒40° +∠BOE=70 °

⇒∠BOE=30°

∠AOE+∠BOE=180 °

[Linear pair of angles]

⇒∠AOE+30 °=180 °

⇒∠AOE=150 °

⇒∠AOC+∠COE=150 °

⇒40 °+∠COE=150 °

⇒∠COE=110 °

Answer 2

Question :-

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

Answer :-

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°.

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Hope it helpsss :)