Question

EXERCISE 0.1
1. In Fig. 6.13, lines AB and CD intersect at O. If/AOC + /BOE = 70° and /BOD = 40°, find
BOB and reflex 2 COE.
A
A
B
D
Fig. 6.13​

Answer 1

Answer:

∠BOE =30° and ∠COE = 110°

Step-by-step explanation:

(sum of angle in linear pair always equal to 180 )

∠AOC + ∠COE + ∠BOE = 180°

Given that ∠ AOC + ∠ BOE = 70° plug this value we get

= > 70° + ∠COE = 180°

= > ∠ COE = 180° -70°

= > ∠ COE = 110°

Also

(sum of angle in linear pair always equal to 180° )

∠COE + ∠BOE + ∠BOD = 180°

Put ∠COE = 110° and ∠ BOD = 40° we get

110° + ∠BOE + 40° = 180°

= > ∠BOE = 180° - 110° - 40°

= > ∠BOE = 30°

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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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