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In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of *x* and *y*.

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#### Solution

It is given that PQ || SR and QR is a transversal line.

∠PQR = ∠QRT (Alternate interior angles)

*x* + 28º = 65º

*x *= 65º − 28º

*x* = 37º

By using the angle sum property for ΔSPQ, we obtain

∠SPQ + *x* + *y* = 180º

90º + 37º + *y* = 180º

*y* = 180º − 127º

*y *= 53º

*∴x* = 37º and *y* = 53º

Concept: Angle Sum Property of a Triangle

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