In a circle with centre O, a diameter AB is produced to a point P lying outside the circle and PT is a tangent to the circle at a point C on it. If ∠BPT = 28°, then what is the measure of ∠BCP ?

31°

OC is perpendicular to PT at C

So, \(\Delta \)OCP is a right angled triangle at C.

Then, \(\angle\)COP = 90 - 28 = 62

Here, OA = OB = Radii of the circle

So, \(\Delta \)OBC is an isosceles triangle.

Then, \(\angle\)BCO = \(\angle\)CBO = (180 - 62)/2 = 59

Here, ABP is a straight line.

So, \(\angle\)CBP = 189 - 59 = 121

Here, \(\Delta \)BCP is triangle.

So,

\(\angle\)BCP + \(\angle\)CBP + \(\angle\)BPC = 180

= \(\angle\)BCP = 180 - 28 - 121

= \(\angle\)BCP = 31

Therefore, \(\angle\)BCP is \({31}^\circ\).