The tangents PK and PM, M and K tangency points are drawn to the circle. Find the angle PMK if the angle MPK = 80 °

Let’s construct the radii OK and OM to the points of tangency K and M.

The radii drawn to the points of tangency are perpendicular to the tangents themselves, then the angle РKO = РMO = 90, and the triangles РKO and РMO are rectangular.

In right-angled triangles RKO and RMO, the hypotenuse RO is common, leg OM = OK = R, then the right-angled triangles РKO and РMO are equal in leg and hypotenuse, and then the angle OPM = OPK and therefore РO is the bisector of the angle KРM. Then the angle OРM = OРK = 80/2 = 40.

The KPM triangle is isosceles, since PM = RK as tangents drawn from one point, then the PH is the bisector, the height and median of the KPM triangle.

In a right-angled triangle, the sum of the acute angles is 90, then the angle РМН = РМК = (90 – 40) = 50.

Answer: The RCC angle is 50.