Two circles, centred at P and Q, intersect at two points C and D. AB is tangent to the two circles at A and B. If ∠ADB = 68°, find the value of ∠ACB.
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Let us analyze the problem step by step:

1. AB is a common tangent to both circles at points A and B. The circles intersect at points C and D.
2. ∠ADB = 68°. We are to find ∠ACB.

Let us consider the circle with center P and tangent at A. Since AB is tangent at A, and D is a point on the circle, ∠DAB = angle in the alternate segment = ∠DCB (by the tangent-secant theorem).

But the key is to use the property of intersecting circles and the fact that AB is tangent to both circles at A and B.

Let us focus on quadrilateral ACBD:
- Points A, C, D, B lie on the circumference of the two circles.
- Since AB is tangent at A and B, and C, D are intersection points, quadrilateral ACBD is cyclic.

In a cyclic quadrilateral, the sum of opposite angles is 180°.
So, ∠ADB + ∠ACB = 180°
Given ∠ADB = 68°
Therefore, ∠ACB = 180° - 68° = 112°

Thus, the answer is 112°.

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