In the given figure 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 that PQ is a tangent at point C to a circle with center O, and AB is a diameter with ∠CAB = 30°, we need to find ∠PCA.

To find ∠PCA, we can use the properties of tangents and angles in a circle.

Step 1: Draw a diagram representing the given information. Label the points as described: PQ is the tangent at point C, AB is the diameter, and O is the center of the circle.

Step 2: Since AB is a diameter, it passes through the center O. Therefore, ∠CAB is an inscribed angle that intercepts the same arc as ∠COB.

Step 3: By the inscribed angle theorem, we know that the measure of an inscribed angle is equal to half the measure of its intercepted arc. Therefore, ∠CAB = 30° implies that the intercepted arc COB has a measure of 60°.

Step 4: Since PQ is a tangent to the circle at point C, it is perpendicular to the radius OC. Therefore, ∠PCO is a right angle.

Step 5: Since ∠PCO is a right angle and ∠COB is 60°, we can find ∠PCA by subtracting the measures of ∠PCO and ∠COB from 180° (the sum of angles in a triangle).

Step 6: ∠PCA = 180° - ∠PCO - ∠COB = 180° - 90° - 60° = 30°.

Therefore, ∠PCA is equal to 30°.