Question

EXERCISE 61
1. In Fig. 6.13. lines AB and CD intersect at on
AOC +2 BOE -70and Z BOD - 40" find
BOE and reflex ZCOE
Fig. 6.13​

Answer 1

Answer:

angle AOC+angleBOE =70°

angle BOD=40°

here,angle AOC =angle BOD(v.o.a)

angle AOC=40°

from equation (1), 40°+angle BOE=70

angle BOE=70°-40℃

=30°

now,CD is a line

angleCoE+angle BOE+ angle BOD=180°

angleCOE+30°+40°=180°

angle COE =180°-70°=110°

reflex angle COE=360°-110℃

reflex angle COE=250°

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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