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If tangents PA and PB from a point P to a circle with center O are drawn so that ∠APB = 80° then ∠POA = ?
In △AOP and △BOP
AP = BP
[Tangents drawn from an external point are equal]
OP = OP [Common]
OA = OB
[Radii of same circle]
△AOP ≅ △BOP
[By Side - Side - Side criterion]
∠APO = ∠BPO
[Corresponding parts of congruent triangles are congruent]
∠APB = ∠APO + ∠BPO
80 = ∠APO + ∠APO
2∠APO = 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°