Two circles touch each other externally at P. AB is a direct common tangent whose touch points are A and B. If (\angle PAB = 35^o), then (\angle ABP) is -

In triangle PAB, PA and PB are radii to points of tangency. A radius is perpendicular to the tangent at the point of contact, so angle OAP = 90° where O is the center of circle 1. This doesn't directly help us. Instead, consider that angle PAB = 35° is given. At point A, the tangent AB is perpendicular to radius OA. Similarly, at point B, tangent AB is perpendicular to radius PB of the second circle. Looking at triangle PAB, we can use the fact that the sum of angles is 180°. Through geometric properties of the tangent configuration, angle BPA = 70°, making triangle PAB isosceles. Therefore angle ABP = (180° - 70°)/2 = 55°.