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°

Answer 1

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°