Question

In figure, the tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA and ∠BCO.

[Hint: Join CO.]

Answer 1

Given:

AB is a diameter of the circle, PC is tangent at C, and ∠PCA = 110°.

Join CO.

Step-wise calculation:

1. Because AB is a diameter, angle in a semicircle gives ∠ACB = 90°.

2. ∠PCA = ∠PCB + ∠BCA

So 110° = ∠PCB + 90°

⇒ ∠PCB = 20°

3. Radius OC is perpendicular to tangent PC.

So ∠OCP = 90°. 

Hence ∠OCB + ∠PCB = 90°

⇒ ∠OCB = 90° – 20°

⇒ ∠OCB = 70°

4. In ΔOCB, OC = OB (radii)

⇒ ∠OBC = ∠OCB = 70°

But ∠OBC is the same as ∠CBA.

∠CBA = 70°

∠BCO = ∠OCB = 70°