The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.

Calculate:

(i) ∠POS,

(ii) ∠QOR,

(iii) ∠PQR.

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

∴ ∠POQ = ∠QOR = ∠ROS = x (let) [Equal chords subtends equal angles at the centre]

Arc PQRS subtends ∠POS at the centre and ∠PTS at the circumference of the circle.

Thus,

∠POS = 2 x ∠PTS = 2 x 75° = 150° [As angle subtended at the centre by the arc is double that it subtends at any point on the circumference of the circle.]

⇒ ∠POQ + ∠QOR + ∠ROS = 150°

⇒ x + x + x = 150°

⇒ 3x = 150°

⇒ x = $\dfrac{150°}{3}$ = 50°.

In ∆OPQ we have,

⇒ OP = OQ [Radii of the same circle]

⇒ ∠OPQ = ∠OQP = y (let) [Angles opposite to equal sides are equal]

In ∆OPQ,

⇒ ∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ + ∠OQP + 50° = 180°

⇒ ∠OPQ + ∠OQP = 180° - 50°

⇒ ∠OPQ + ∠OQP = 130°

⇒ 2y = 130°

⇒ y = $\dfrac{130°}{2}$ = 65°

⇒ ∠OPQ = ∠OQP = y = 65°.

In ∆OQR we have,

⇒ OQ = OR [Radii of the same circle]

⇒ ∠OQR = ∠ORQ = z (let) [Angles opposite to equal sides are equal]

In ∆OQR

⇒ ∠OQR + ∠ORQ + ∠QOR = 180°

⇒ z + z + 50° = 180°

⇒ 2z = 180° - 50°

⇒ 2z = 130°

⇒ z = $\dfrac{130°}{2}$ = 65°

⇒ ∠OQR = ∠ORQ = z = 65°.

(i) Hence, ∠POS = 150°.

(ii) Hence, ∠QOR = 50°.

(iii) From figure,

∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°.

Hence, ∠PQR = 130°.