Question

EXERCISE 6.1
1. In Fig. 6.13, lines AB and CD intersect at O. If
Z AOC + Z BOE = 70° and Z BOD = 40°, find
BOE and reflex 2 COE.
с C
A
D'S
Fig. 6.13​

Answer 1

Step-by-step explanation:

Given that,

AB is a straight line. Lines AB and CD intersect at O. \angle AOC + \angle BOE = 70^0 and \angle BOD = 40^0

Since AB is a straight line

\therefore \angleAOC + \angleCOE + \angleEOB = 180^0

\Rightarrow \angle COE = 180^0-70^0=110^0 [since \angle AOC + \angle BOE = 70^0]

So, reflex \angleCOE = 360^0-110^0 = 250^0

It is given that AB and CD intersect at O

Therefore, \angleAOC = \angleBOD [vertically opposite angle]

\Rightarrow \angle COA = 40^0 [ GIven \angle BOD = 40^0]

Also, \angle AOC + \angle BOE = 70^0

So, \angleBOE = 30^0

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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