n 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.

In the given figure, we have a circle with center O. PQ is a tangent to the circle at point C. AB is a diameter of the circle. We are given that ∠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 line segment from O to C, and label it as OC. Since OC is a radius of the circle, it is perpendicular to PQ.

Step 2: Since AB is a diameter, ∠CAB is an inscribed angle that intercepts the same arc as ∠COB. By the inscribed angle theorem, we know that the measure of an inscribed angle is half the measure of its intercepted arc. Therefore, ∠CAB = 30° implies that the intercepted arc COB has a measure of 60°.

Step 3: Since OC is perpendicular to PQ, ∠OCB is a right angle. Therefore, ∠COB = 90°.

Step 4: Using the fact that the sum of angles in a triangle is 180°, we can find ∠CBO. ∠CBO = 180° - ∠COB - ∠OCB = 180° - 90° - 60° = 30°.

Step 5: Since ∠CBO = ∠CAB, we have a pair of congruent angles. Therefore, triangle CBO is an isosceles triangle.

Step 6: In an isosceles triangle, the angles opposite the congruent sides are also congruent. Therefore, ∠CBO = ∠BCO = 30°.

Step 7: Since ∠BCO is an exterior angle of triangle PCA, we can use the exterior angle theorem to find ∠PCA. ∠PCA = ∠BCO + ∠CBO = 30° + 30° = 60°.

Therefore, ∠PCA = 60°.