In a circle with center 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?
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Correct answer: D
Explanation
The correct answer is option 4 i.e. 31°
PT is a tangent to the circle at point C.
∠BPT = 28o
In Δ OCP -
∠OCP + ∠COP + ∠OPT = 180o
90o + ∠COP + 28o = 180o (OC is perpendicular to tangent PT)
∠COP = 180o - 118o = 62o
In Δ OCB -
Let ∠OCB = ∠OBC = x (OC = OB = Radius)
∠OCB + ∠OBC + ∠COB = 180o
x + x + 62o = 180o
2x = 118o
x = 59o
∠OCB = ∠OBC = x = 59o
∠OCP = ∠OCB + ∠BCP
90o = 59o + ∠BCP
∠BCP = 90o - 59o = 31o