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Question
A, B, and C are three points on a circle. The tangent at C meets BN produced at T. Given that ∠ ATC = 36° and ∠ ACT = 48°, calculate the angle subtended by AB at the center of the circle.
Solution
Join BC. Let O be the centre of a circle. Join OA and OB
In Δ BCT, Δ ACT,
∠BTC = ∠ATC = 36°
∠ACT = ∠ABC = 48°
∴∠BAC = ∠ACT + ∠ATC
∠BAC = 48° + 36° = 84°
∴∠BCA = 180° - (∠ABC +∠BAC )
∠BCA = 180° - (48° + 84°) = 48°
∴∠BOA = 2∠BCA
∠BOA = 2 x 48° = 96°
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