In the adjoining figure, three coplanar lines AB, CD and EF intersect

1.	In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence, find ∠AOD, ∠COE and ∠AOE.
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 above 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 On further calculation 10x = 180^o By division x = 180/10 = 18 By substituting the value of x ∠AOD = 2x^o So we get ∠AOD = 2 (18)^o = 36^o ∠EOA = ∠BOF = 3x^o So we get ∠EOA = ∠BOF = 3 (18)^o = 54^o ∠COE = 5x^o So we get ∠COE = 5 (18)^o = 90^o

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence, find ∠AOD, ∠COE and ∠AOE.

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 above 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

On further calculation

10x = 180^o

By division

x = 180/10 = 18

By substituting the value of x

∠AOD = 2x^o

So we get

∠AOD = 2 (18)^o = 36^o

∠EOA = ∠BOF = 3x^o

So we get

∠EOA = ∠BOF = 3 (18)^o = 54^o

∠COE = 5x^o

So we get

∠COE = 5 (18)^o = 90^o