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Solution:
From the figure we know that ∠AOD and ∠BOC are vertically opposite angles
∠AOD = ∠BOC
It is given that
0
∠BOC + ∠AOD = 280
We know that ∠AOD = ∠BOC
∠AOD + ∠AOD = 280
0
2 ∠AOD = 280
∠AOD = 280/2
0
∠AOD = ∠BOC = 140 we know that ∠AOC and ∠AOD form a linear pair.
0
∠AOC + ∠AOD = 180
0
∠AOC + 140 = 180
0
0
∠AOC = 180 - 140
0
∠AOC = 40
From the figure we know that ∠AOC and ∠BOD are vertically opposite angles
0
∠AOC = ∠BOD = 40
0
0
0
0
Therefore, ∠AOC = 40 , ∠BOC = 140 , ∠AOD = 140 and ∠BOD = 40

13. In the given figure, the two lines AB and CD intersect at a point ‘O’ such that
0
∠BOC = 125 . Find the values of x, y and z.

Solution:
From the figure we know that ∠AOC and ∠BOC form a linear pair of angles.
0
∠AOC + ∠BOC = 180
0
x + 125 = 180
0
0
x = 180 - 125
0
x = 55
From the figure we know that ∠AOD and ∠BOC are vertically opposite angles
So, we get
0
y = 125
From the figure we know that ∠BOD and ∠AOC are vertically opposite angles
So, we get
0
z = 55
0
0
0
Therefore, the values of x, y and z are 55 , 125 and 55 .

14. If two straight lines intersect each other than prove that the ray opposite to
the bisector of one of the angles so formed bisects the vertically opposite
angle.
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