# If tangents PA and PB from a point P to a circle with center O are drawn so that ∠APB = 80° then ∠POA = ?A. 40°B. 50°C. 80°D. 60°

In AOP and BOP

AP = BP

[Tangents drawn from an external point are equal]

OP = OP [Common]

OA = OB

AOP BOP

[By Side - Side - Side criterion]

APO = BPO

[Corresponding parts of congruent triangles are congruent]

APB = APO + BPO

80 = APO + APO

2APO = 80

APO = 40°

In AOP

APO + AOP + OAP = 180°

[By angle sum property]

40° + AOP + 90° = 180°

[ OAP = 90° as OA AP because Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

AOP = 50°

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