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