In the figure (i) given below, PQ is a diameter. Chord SR is parallel to PQ. Given ∠PQR = 58°, calculate

(i) ∠RPQ

(ii) ∠STP

(T is a point on the minor arc SP)

(i) From figure,

∠PRQ = 90° (∵ angle in semicircle is equal to 90°.)

Since sum of angles in triangle is 180°.

⇒ ∠PRQ + ∠RQP + ∠RPQ = 180°
⇒ 90° + 58° + ∠RPQ = 180°
⇒ ∠RPQ + 148° = 180°
⇒ ∠RPQ = 180° - 148
⇒ ∠RPQ = 32°.

Hence, the value of ∠RPQ = 32°.

(ii) From figure,

∠SRP = ∠RPQ = 32° (∵ alternate angles are equal.)

Since, PTSR is a cyclic quadrilateral so sum of its opposite angles is equal to 180°.

⇒ ∠SRP + ∠STP = 180°
⇒ 32° + ∠STP = 180°
⇒ ∠STP = 180° - 32°
⇒ ∠STP = 148°.

Hence, the value of ∠STP = 148°.