Question

EXERCISE 1
1. In Fig. 6.13. lines AB and CD intersect at O. If
AOC+BOF= 70° and BOD = 40°, find
BOH and refler_COE
Fig. 6.13​

Answer 1

Step-by-step explanation

Answers

Aoc + boe = 70 given

bod = 40

aoc = bod =40 (vertically opposite angle)

so

aoc + boe = 70

boe = 70-40=30

aob = 180 ( linear pair)

aoc+ coe+ boe = 180

40 + coe +30 =180

coe=180-70=110

now reflex coe

aoc+ aod+bod+boe=360

aod= coe( vertically opposite angle)

reflex coe= 360

coe= 110

reflex coe = 360-coe

360-110=250 .....!!!!!!

hope it help you

this is ncert class 9 question in chapter 6

exercise 6.1 question no.1

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°

o r (∠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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