Page 9 - Ans Sub

11. In the given figure, AB || PQ. Find the values of x and y.

Solution:
It is given that AB || PQ and EF is a transversal
From the figure we know that ∠CEB and ∠EFQ are corresponding angles
So, we get
0
∠CEB = ∠EFQ = 75
It can be written as
0
∠EFQ = 75
Where
0
∠EFG + ∠GFQ = 75
0
0
0
25 + y = 75
0
0
0
y = 75 - 25
0
0
y = 50
From the figure we know that ∠BEF and ∠EFQ are co-interior angles.
So, we get
0
∠BEF + ∠EFQ = 180
0
0
∠BEF + 75 = 180
0
0
∠BEF = 180 - 75
0
∠BEF = 105
We know that ∠BEF can be written as
∠BEF = ∠FEG + ∠GEB
0
0
105 = ∠FEG + 20
0
0
∠FEG = 105 - 20
0
∠FEG = 85
Now, in △ EFG,
0
0
0
x + 25 + ∠FEG = 180
0
0
0
0
x + 25 + 85 = 180
0
0
0
0
x = 180 - 25 - 85
0
0
0
x = 180 – 110
0
0
x = 70
0
Therefore, the value of x is 70 .

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