In the given figure, PQ and PR are tangents to a circle with centre A.

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).