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 the point C on it. If ∠BPT = 36º, then what is the measure of ∠BCP?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 27º
Concept
Chord and Tangent intersect to create a 90º
Application
In triangle OCP
∠OCP + ∠COP + ∠OPT = 180°
⇒ 90° + ∠COP + 36° = 180°
∠COP = 180° - 126° = 54°
OB = OC (Radius of circle)
In triangle OCB
∠OCB + ∠OBC + ∠COB = 180°
Let ∠OCB = ∠OBC = K°
According to the question
K + K + 54 = 180
2K = 126°
K = 63°
Chord is perperdicular to the tangent PT
∠OCP = 90°
∠OCP = ∠OCB + ∠BCP
90° = 63° + ∠BCP
∠BCP = 90° - 63° = 27°