In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

In the given figure,

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°