Question

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.

Answer 1

In the given figure,
In ΔACO,

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°