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Question
A tangent ADB is drawn to a circle at D whose centre is C. Also, PQ is a chord parallel to AB and ∠QDB = 50°. Find the value of ∠PDQ.
Solution
Given: Chord PQ || tangent AB
∴ ∠Q = ∠QDB = 50° ......[Alternate interior angles]
And ∠CDB = ∠PRD ......[Alternate interior angles]
ADB is a tangent at point D and CD is the circle's radius.
Thus, ∠CDB = 90° ......[Tangent theorem]
⇒ ∠CDQ + ∠QDB = 90°
⇒ ∠CDQ + 50° = 90°
⇒ ∠CDQ = 90° – 50° = 40°
CR is a perpendicular to the chord PQ, so it bisects PQ.
Thus, ∠PRC = 90° and PR = QR
Now, in ΔDRP and ΔDRQ
PR ≅ QR .....[Proved above]
∠DRP ≅ ∠DRQ ......[Each 90°]
DR ≅ DR ......[Common side]
Thus ΔDRP ≅ ΔDRQ .....[By SAS criterion of congruence)
So, ∠P ≅ ∠Q .....[C.A.C.T.]
⇒ ∠P = 50° .....[∵ ∠Q = 50°]
Now, the sum of the interior angles in a triangle is 180°.
Thus, ∠P + ∠PRD + ∠RDP = 180°
50° + 90° + ∠RDP = 180°
∠RDP = 180° – 50° – 90° = 40°
Now, ∠PDQ = ∠RDP + ∠RDQ
⇒ ∠PDQ = 40° + 40° = 80°
Hence, ∠PDQ = 80°.
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