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If PA and PB are two tangents to a circle with center O such that ∠AOB = 110° then ∠APB is equal to
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + 110° + ∠APB = 360°
∠APB = 70°