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Question
In the given figure, AB and CD are parallel lines and transversal EF intersects them at Pand Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
Options
x = 55°, y = 40°
x = 50°, y = 45°
x = 60°, y = 35°
x = 35°, y = 60°
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Solution
In the given figure and AB|| CD , ∠APR = 25°, ∠RQC = 30V , and ∠CQF = 65°
We need to find the value of x and y
Here, we draw a line ST parallel to AB, i.e AB || ST
Also, using the property, “two lines parallel to the same line are parallel to each other”
As,
AB || ST
AB || CD
Thus, CD || ST
Now, AB || ST and EF is the transversal, so using the property, ”alternate interior angles are equal”, we get,
∠APR = ∠PRT
∠PRT = 25° .......... (1)
Similarly, CD || ST and EF is the transversal
∠QRT = ∠RQC
∠QRT = 30° .......(2)
Adding (1) and (2), we get
∠PRT + ∠QRT = 25° + 30°
x = 55°
Further,FPE is a straight line
Applying the property, angles forming a linear pair are supplementary
∠CQF + ∠CQR + ∠RQP = 180°
65° + 30° + ∠RQP = 180°
∠RQP = 180° - 95°
∠RQP = 85°
Also, applying angle sum property of the triangle
In ΔPRQ
∠RPQ + 55° + 85° = 180°°
140° + y = 180°
y = 180° - 140°
y = 40°
Thus, x = 55° and y = 40°
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