In the given figure, find ∠BDC

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Solution

∠BOC is a central angle=150^o(given)
Angle at centre is double the angle on remaining part of circle
∠BAC=75^o
In a cyclic quadrilateral ACDB
∠BAC+∠BDC=180^o
∠BDC=180^o−∠BAC
$∠ B D C = 105^{\circ}$

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