Page 5 - ANS SUB

0
0
0
0
90 + x + 50 = 180
0
0
0
x + 140 = 180
0
0
0
x = 180 – 140
0
0
x = 40 ∠EOB and ∠AOF are vertically opposite angles.
0
∠EOB = ∠AOF = x = y = 40
0 ,
0,
0 ,
0
Therefore, the values of x, y, z and t are 40 40 50 90 .
11. 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.

Solution:
From the figure we know that ∠COE and ∠EOD form a linear pair
0
∠COE + ∠EOD = 180
0
∠COE + ∠EOA + ∠AOD = 180
0
5x + ∠EOA + 2x = 180 ∠EOA and ∠BOF are vertically opposite angles.
∠EOA = ∠BOF
0
5x + ∠BOF + 2x = 180
5x + 3x + 2x = 180
0
10x = 180
x = 180/10 = 18
0
∠AOD = 2x
0
0
∠AOD = 2 (18) = 36
0
∠EOA = ∠BOF = 3x
0
0
∠EOA = ∠BOF = 3 (18) = 54
0
∠COE = 5x
0
0
∠COE = 5 (18) = 90

0
12. Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280 ,
as shown in the
Figure. Find all the four angles.

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