#### Question

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

(i) ∠POS, (ii) ∠QOR, (iii) ∠PQR.

#### Solution

Join OP, OQ and OS.

∵ PQ = QR = RS,

∠POQ = ∠QOR = ∠ROS

[Equal chords subtends equal angles at the centre]

Arc PQRS subtends ∠POS at the center and ∠PTS at the remaining Parts of the circle.

∴ ∠POS = 2∠PTS = 2 × 75° = 150°

⇒ ∠POQ + ∠QOR + ∠ROS = 150°

⇒ `∠POQ = ∠QOR = ∠ROS = (150° )/3 = 50° `

In Δ OPQ, OP = OQ [radii of the same circle]

∴ ∠OPQ = ∠OQP

But ∠OQP + ∠POQ = 180°

∴ ∠OPQ + ∠QP = 50° = 180°

⇒ ∠OPQ + ∠OQP = 180° - 50°

⇒ ∠OPQ + ∠OPQ = 130°

⇒ 2∠OPQ = 130°

⇒ `∠OPQ = ∠OQP = (130°)/2 = 65° `

Similarly, we can prove that

In ΔOQR, ∠OQR = ∠ORQ = 65°

And in ΔORS, ∠ORS = OSR = 65°

(i) Now ∠POS = 150°

(ii) QOR = 50° and

(iii) ∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°