We are looking for BCA:
OBD = 90
Reason: Line BD is tangent to circle, and as such it forms a 90 degrees with point of tangency.
Similarly: OAD = 90
Reason: Line DA is tangent to circle, and as such it forms a 90 degrees with point of tangency.
So angles in quadrilateral: BOAD
56, 90, 90 and BOA
56 + 90 + 90 + BOA = 360
BOA = 360 - 90 -90 - 56
BOA = 124
Angle BCA is angle at circumference
Angle BOA = 124, is the angle at centre
Angle at Centre = 2* Angle at circumference
Angle at Circumference = Angle at Centre / 2 = 124/2 = 62
Angle BCA = 62 degrees
Option B.
I hope this helps.
