Question

In the given figure, find ∠BDC

Answer 1

∠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
$\angle BDC={105}^{\circ }$