Q. 225.0( 1 Vote )

In the given figure, PQ and PR are tangents to a circle with center A. If ∠QPA = 27° then ∠QAR equalsA. 63°B. 117°C. 126°D. 153°

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°

AQP + ARP + QAR + QPR = 360°

90° + 90° + QAR + 54° = 360°

QAR = 126°

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