Question

In the figure pq is the tangent to a circle with center o at the point of contact c if AB is a diameter and

Answer 1

In Δ ACO,

OA=OC (Radii of the same circle)

Therefore,

ΔACO is an isosceles triangle.

∠CAB = 30° (Given)

∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)

∠PCO = 90° …(radius is drawn at the point of

contact is perpendicular to the tangent)

Now ∠PCA = ∠PCO – ∠CAO

Therefore,

∠PCA = 90° – 30° = 60°