In Fig. 1, PQ is a tangent at a point C to a circle with centre O. if AB is a diameter and ∠CAB = 30°, find ∠PCA.
In the given figure,
OA = OC …(Radii of the same circle)
∴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 drawn at the point of contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
∴∠PCA = 90° – 30° = 60°