Two circles touch each other externally at C and AB is common tangent to the circles then $\angle ACB$ is :

Let the common tangent AB touch the first circle at A and the second at B, and let the centres be O and P. The radii OA and PB are both perpendicular to the tangent AB, making OA parallel to PB. Drawing the line segments AC and BC forms right triangles OCA and PCB, where OC = AC and PC = BC because they are tangents from an external point to a circle. Since angle OAC + angle PBC = 180 degrees, the quadrilateral OABP is a trapezoid, and dropping perpendiculars shows that angle OCA = angle CAB and angle PCB = angle CBA. The sum of angles CAB and CBA is 90 degrees, making angle ACB equal to 90 degrees.

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