Question

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

Calculate:

1. ∠POS, 
2. ∠QOR, 
3. ∠PQR.

Answer 1

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^circ)/3 = 50^circ`

In ΔOPQ, OP = OQ   ...[Radii of the same circle]

∴ ∠OPQ = ∠OQP         

But ∠OPQ + ∠OQP + ∠POQ = 180° 

∴ ∠OPQ + ∠QP = 50° = 180°

`=>` ∠OPQ + ∠OQP = 180°  – 50°

`=>` ∠OPQ + ∠OPQ = 130°

`=>` 2∠OPQ = 130°

`=> ∠OPQ = ∠OQP = (130^circ)/2 = 65^circ`

Similarly, we can prove that

In ΔOQR, ∠OQR = ∠ORQ = 65°

And in ΔORS, ∠ORS = OSR = 65°

1. Now ∠POS = 150°
2. QOR = 50° and
3. ∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°