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Sum of Opposite Angles of a Cyclic Quadrilateral

In the given ...

Question

In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ∠PSR = 150°, find ∠RPQ.

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Solution

In cyclic quadrilateral PQRS, we have:
∠PSR + ∠PQR = 180°
⇒ 150° + ∠PQR = 180°
⇒ ∠PQR = (180° – 150°) = 30°
∴ ∠PQR = 30° ...(i)
Also, ∠PRQ = 90° (Angle in a semicircle) ...(ii)
Now, in ΔPRQ, we have:
∠PQR + ∠PRQ + ∠RPQ = 180°
⇒ 30° + 90° + ∠RPQ = 180° [From(i) and (ii)]
⇒ ∠RPQ = 180° – 120° = 60°
∴ ∠RPQ = 60°

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