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प्रश्न
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
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उत्तर
Join PR.
i. ∠PRQ = 90°
(Angle in a semicircle)
∴ In right triangle PQR,
∠RPQ = 90° – ∠PQR
= 90° – 58°
= 32°
ii. Also, SR || PQ
∠PRS = ∠RPQ = 32° (Alternate angles)
In cyclic quadrilateral PRST,
∠STP = 180° – ∠PRS
= 180° – 32°
= 148°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
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