Mathematics
In the following figure, PQ and PR are tangents to the circle, with center O. If ∠QPR = 60°, calculate :
(i) ∠QOR,
(ii) ∠OQR,
(iii) ∠QSR.
Circles
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Answer
(i) We know that,
The tangent at any point of a circle and the radius through this point are perpendicular to each other.
In quadrilateral ORPQ,
∠OQP = ∠ORP = 90° [∵ The tangent at any point of a circle and the radius through this point are perpendicular to each other]
∠QPR = 60° [Given]
⇒ ∠OQP + ∠ORP + ∠QPR + ∠QOR = 360° [By angle sum property of quadrilateral]
⇒ 90° + 90° + 60° + ∠QOR = 360°
⇒ 240° + ∠QOR = 360°
⇒ ∠QOR = 120°.
Hence, ∠QOR = 120°.
(ii) Join QR.
In △QOR,
OQ = OR (Radii of same circle)
As, angles opposite to equal sides are equal.
∴ ∠OQR = ∠ORQ ……….(1)
⇒ ∠OQR + ∠ORQ + ∠QOR = 180°
⇒ ∠OQR + ∠ORQ + 120° = 180°
⇒ ∠OQR + ∠ORQ = 180° - 120°
⇒ ∠OQR + ∠ORQ = 60°
⇒ ∠OQR + ∠OQR = 60°
⇒ 2∠OQR = 60°
⇒ ∠OQR =
⇒ ∠OQR = 30°.
Hence, ∠OQR = 30°.
(iii) Arc RQ subtends ∠QOR at the center and ∠QSR at the remaining part of the circle.
We know that,
The angle subtended by an arc at the centre is twice the angle subtended at the circumference.
⇒ ∠QOR = 2∠QSR
⇒ ∠QSR = ∠QOR = = 60°.
Hence, ∠QSR = 60°.
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