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Question
In the following figure, PQ and PR are tangents to the circle, with centre O. If `∠`QPR = 60°, calculate:
- ∠QOR,
- ∠OQR,
- ∠QSR.
Solution
Join QR.
i. In quadrilateral ORPQ,
OQ ⊥ OP, OR ⊥ RP
∴ ∠OQP = 90°, ∠ORP = 90°, ∠QPR = 60°
∠QOR = 360° – (90° + 90° + 60°)
∠QOR = 360° – 240°
∠QOR = 120°
ii. In ΔQOR,
OQ = QR ...(Radii of the same circle)
∴ ∠OQR = ∠QRO ...(i)
But, ∠OQR + ∠QRO + ∠QOR = 180°
∠OQR + ∠ QRO + 120° = 180°
∠OQR + ∠QRO = 60°
From (i)
2∠OQR = 60°
∠OQR = 30°
iii. Now arc RQ subtends ∠QOR at the centre and ∠QSR at the remaining part of the circle.
∴ `∠QSR = 1/2 ∠QOR`
`=> ∠QSR = 1/2 xx 120^circ`
`=>` ∠QSR = 60°
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