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Question -

From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE 2016]



Answer -


PA = PB [tangents drawn from external point are equal]
∠PBA = ∠PAB = 50° [angles equal to opposite sides]
∠APB = 180° – 50° – 50° = 80° [angle-sum property of a A]
In cyclic quad. OAPB
∠AOB + ∠APB = 180° [sum of opposite angles of a cyclic quadrilateral is 180°]
∠AOB + 80° = 180°
∠AOB = 180°- 80° = 100°

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