Question

In Fig. 20, OE is the bisector of ZBOD. If <1 = 70°, find the magnitudes of 2 ,3 and 4​

Answer 1

Answer:

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Step-by-step explanation:

Let ∠EOB=x

Given,

∠EOB=∠EOD [ ∵ OE is bisector of ∠BOD]

x=∠1.

=> ∠1=x=70°

=> ∠BOD=∠1+x=(70+70)°

=>∠BOD=140°

According to the figure,

∠COB=∠AOD and

∠AOC=∠BOD=140 °

[ ∵ vertically opposite angles]

We know that,

sum of all the angles around a point =360°

∠BOD+∠AOD+∠AOC+∠COB=360 °

∠BOD+∠AOD+∠BOD+∠AOD=360°

2∠BOD+2∠AOD=360°

∠BOD+∠AOD=180°

140+∠AOD=180°

∠AOD=180°

−140°

∠AOD=20°

=> ∠COB=20°

∠AOC=∠BOD=140°

∴∠2=∠4=20°

and ∠3=140°