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.

Answer 1

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°.