Mathematics
In the following figure, PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with center O.
Calculate the values of :
(i) ∠QOP
(ii) ∠QCP
Circles
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Answer
(i) Given,
PQ = QR
∴ ∠PRQ = ∠QPR [Angles opposite to equal sides are equal in a triangle.]
In △PQR,
⇒ ∠PRQ + ∠QPR + ∠RQP = 180°
⇒ ∠PRQ + ∠PRQ + 68° = 180°
⇒ 2∠PRQ = 180° - 68°
⇒ 2∠PRQ = 112°
⇒ ∠PRQ =
⇒ ∠PRQ = 56°.
We know that,
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠QOP = 2∠PRQ = 2 x 56 = 112°.
Hence, ∠QOP = 112°.
(ii) We know that,
A tangent line is always at a right angle to the radius of the circle at the point of tangency.
∴ ∠OPC = 90° and ∠OQC = 90°.
In quadrilateral OQCP,
⇒ ∠QOP + ∠OPC + ∠OQC + ∠QCP = 360° [Angle sum property of quadrilateral]
⇒ 112° + 90° + 90° + ∠QCP = 360°
⇒ ∠QCP = 360° - 292° = 68°.
Hence, ∠QCP = 68°.
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