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# In Figure, if PQ ⊥ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of x and y.

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In Figure, if PQ ⊥ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of x and y.

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x + SQR = QRT (As they are alternate angles since QR is transversal)

x + 28° = 65°

∴ x = 37°

It is also known that alternate interior angles are same and so,

QSR = x = 37°

QRS + QRT = 180° (As they are a Linear pair)

QRS + 65° = 180°

QRS = 115°

Now, we know that the sum of the angles in a quadrilateral is 360°

P + Q + R + S = 360°

Putting their respective values, we get,

S = 360° - 90° - 65° - 115°

In Δ SPQ

∠SPQ + x + y = 180°

90° + 37° + y = 180°

y = 180° – 127° = 53°

Hence, y = 53°

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