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प्रश्न
The given figure shows a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.
Calculate:
- ∠POS,
- ∠QOR,
- ∠PQR.
उत्तर
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°
- Now ∠POS = 150°
- QOR = 50° and
- ∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°
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