In Fig. 20, OE is the bisector of angle BOD. If angle 1 = 70°, find the magnitudes of angles 2,3,4.
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Answer:
Let ∠EOB=x
Given,
∠EOB=∠EOD [ ∵ OE is bisector of ∠BOD]
x=∠1.
=> ∠1=x=70
o
.
=> ∠BOD=∠1+x=(70+70)
o
.
=>∠BOD=140
o
.
According to the figure,
∠COB=∠AOD and
∠AOC=∠BOD=140
o
[ ∵ vertically opposite angles]
We know that,
sum of all the angles around a point =360
o
∠BOD+∠AOD+∠AOC+∠COB=360
o
∠BOD+∠AOD+∠BOD+∠AOD=360
o
2∠BOD+2∠AOD=360
o
∠BOD+∠AOD=180
o
140+∠AOD=180
o
∠AOD=180
o
−140
o
∠AOD=20
o
=> ∠COB=20
o
∠AOC=∠BOD=140
o
∴∠2=∠4=20
o
and ∠3=140
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