Question

In the figure, three lines AB, CD and EF intersect at O. Find the
value of ‘x’ and also find angles AOE and BOD.

Answer 1

Answer:

From the figure we know that ∠COE and ∠EOD form a linear pair

∠COE+∠EOD=180

o

It can also be written as

∠COE+∠EOA+∠AOD=180

o

By substituting values in the equation we get

5x+∠EOA+2x=180

o

From the figure we know that ∠EOA and ∠BOF are vertically opposite angles

∠EOA=∠BOF

so we get

5x+∠BOF+2x=180

o

5x+3x+2x=180

o

10x=180

o

x=18

By substituting the value of x

∠AOD=2x

o

∠AOD=2(18)=36

o

∠EOA=∠BOF=3x

o

So we get

∠EOA=∠BOF=3(!8)=54

o

∠COE=5x

o

so we get

∠COE=5(18)=90

o

Step-by-step explanation:

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Answer 2

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It is given that ∠BOD = 180°

from the figure we know that ∠BOD and ∠AOC are vertically opposite angles

∠AOC = ∠BOD = 40°

It is given that ∠AOE = 35°

from the figure we know that ∠BOF and ∠AOE are vertically opposite angles

∠AOE = ∠BOF = 35°

from the figure, we know that AOB is a straight line

so it can be written as

∠AOB = 180°

we can write it as

∠AOE + ∠EOD + ∠BOD = 180°

by substituting the values

35° + ∠EOD + 40°

=180°

∠EOD = 105°

from the figure, we know that ∠COF and ∠EOD are vertically opposite angles

∠COF = ∠EOD = 105°