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Question
In the adjoining figure, PQ is the diameter, chord SR is parallel to PQ. Give ∠ PQR = 58°.
Calculate:
(i) ∠ RPQ,
(ii) ∠ STP ( T is a point on the minor arc)
Solution
(i) ∠ PRQ = 90° ...(∠in a semicircle)
In Δ PQR,
∠ RPQ + ∠ PQR + ∠ PRQ = 180° ....(∵ The sum of the three ∠s of a Δ is 180°)
⇒ ∠ RPQ + 58° + 90° = 180°
⇒ ∠ RPQ + 148° = 180°
⇒ ∠ RPQ = 180° - 148°
⇒ ∠ RPQ = 32°
(ii) ∵ PQ || SR and RP intersects them
∠ PRS = ∠ RPQ ...(Alternate angles)
∴ ∠ PRS = 32°
∵ PTSR is a cyclic quadrilateral.
∴ ∠ PTS + ∠ PRS = 180° ....( ∵ Opposite ∠s of a cyclic quadrilateral are supplementary)
⇒ ∠ PTS + 32° = 180°
⇒ ∠ PTS = 180° - 32° = 148°
⇒ ∠ STP = 148°.
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