Question

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27^∘ then ∠QAR equals [CBSE 2012]
(a) 63^∘
(b) 117^∘
(c) 126^∘
(d) 153^∘

Answer 1

We know that the radius and tangent are perperpendular at their point of contact
Now, In △PQA
∠PQA + ∠QAP + ∠APQ = 180^∘ [Angle sum property of a triangle]
⇒ 90^∘ + ∠QAP + 27^∘ = 180^∘ [∵∠OAB = ∠OBA ]
⇒ ∠QAP = 63^∘
In △PQA and △PRA
PQ = PR (Tangents draw from same external point are equal)
QA = RA (Radii of the circle)
AP = AP (common)
By SSS congruency
△PQA ≅ △PRA
∠QAP = ∠RAP = 63^∘
∴∠QAR = ∠QAP + ∠RAP = 63^∘ + 63^∘ = 126^∘
Hence, the correct answer is option (c).