**Question:**

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27∘ then ∠QAR equals

(a) 63∘

(b) 117∘

(c) 126∘

(d) 153∘

**Solution:**

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).