Question

In the fig. below, if AB || CD, $\angle$APQ = 50${}^{\circ }$ and $\angle$PRD = 127${}^{\circ }$, find x and y.

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• A. x = 50${}^{\circ }$ and y = 77${}^{\circ }$
• B. x = 55${}^{\circ }$ and y = 75${}^{\circ }$
• C. x = 45${}^{\circ }$ and y = 70${}^{\circ }$
• D. x = 55${}^{\circ }$ and y = 65${}^{\circ }$

Answer 1

The correct option is A x = 50${}^{\circ }$ and y = 77${}^{\circ }$

Since AB || CD

∠APQ = ∠PQR (Alternate interior angles of transversal PQ)

⇒ x = 50${}^{\circ }$

∠APR = ∠PRD (Alternate interior angles of transversal PR)

⇒ y + 50${}^{\circ }$ = 127${}^{\circ }$

⇒ y = 127${}^{\circ }$ - 50${}^{\circ }$

⇒ y = 77${}^{\circ }$.