In the given figure, In ∆ACO, OA = OC ... (Radii of the same circle) ∴ ∆ACO is an isosceles triangle. ∠CAB = 30° ... (Given) ∴ ∠CAO = ∠ACO = 30° ... (angles opposite to equal sides of an isosceles triangle are equal) ∠PCO = 90° ... (The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.) Now ∠PCA = ∠PCO – ∠CAO ∴ ∠PCA = 90° – 30° = 60°