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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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