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

Given: PQ is a tangent. AB is a diameter, ∠CAB = 30°
To find: ∠PCA=?
In ΔAOC, ∠CAB = ∠OCA (Angles opposite to equal sides are equal)
So, ∠CAB = 30° = ∠OCA
Since OC ⊥ PQ (Tangent is perpendicular to the radius at point of contact)
∠PCO = 90°
∠OCA + ∠PCA = 90°
30° + ∠PCA = 90°
∠PCA = 90° - 30°
Therefore, ∠PCA = 60°