In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°, Calculate :

(i) ∠RPQ

(ii) ∠STP.

Join PR.

(i) ∠PRQ = 90° [Angle in semi-circle is a right angle.]

In △PQR,

⇒ ∠RPQ + ∠PRQ + ∠PQR = 180° [Angle sum property of triangle]

⇒ ∠RPQ + 90° + 58° = 180°

⇒ ∠RPQ + 148° = 180°

⇒ ∠RPQ = 180° - 148° = 32°.

Hence, ∠RPQ = 32°.

(ii) As, SR || PQ,

∠PRS = ∠RPQ = 32° [Alternate angles are equal]

In cyclic quadrilateral PRST,

⇒ ∠STP + ∠PRS = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠STP = 180° - ∠PRS = 180° - 32° = 148°.

Hence, ∠STP = 148°.