In Given Figure,

PQ = PR…[1]

[Tangents drawn from an external point are equal]

In △AOP and △BOP

PQ = PR [By 1]

AP = AP [Common]

AQ = AR [radii of same circle]

△AQP ≅△ARP [By Side - Side - Side Criterion]

∠QPA = ∠RPA

[Corresponding parts of congruent triangles are congruent]

Now,

∠QPA + ∠RPA = ∠QPR

∠QPA + ∠QPA = ∠QPR

2 ∠QPA = ∠QPR

∠QPR = 2(27) = 54°

As PQ and PQ are tangents to given circle,

We have,

AQ ⏊ PQ and AR ⏊ PR

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠AQP = ∠ARP = 90°

In quadrilateral AQRP, By angle sum property of quadrilateral, we have

∠AQP + ∠ARP + ∠QAR + ∠QPR = 360°

90° + 90° + ∠QAR + 54° = 360°

∠QAR = 126°