In the triangle MPK, a circle is inscribed, O is its center. The angle M is 50 degrees, the angle K is 70 degrees.

In the triangle MPK, a circle is inscribed, O is its center. The angle M is 50 degrees, the angle K is 70 degrees. Calculate the degree measures of the angle MOK, MOP, POK.

Find the degree measure of the angle P:
angle M + angle P + angle K = 180 degrees (according to the theorem on the sum of the angles of a triangle);
50 + angle P + 70 = 180;
angle P = 180 – 120;
angle P = 60 degrees.
From the basic properties of tangents, it is known that segments of tangents to a circle drawn from one point make equal angles with a straight line passing through this point and the center of the circle. That is, the segments OM, OK and OP will be the bisectors of the angles M, K and P, respectively. Then:
angle KMO = angle RMO = angle M / 2 = 50/2 = 25 degrees;
angle MCO = angle RCO = angle K / 2 = 70/2 = 35 degrees;
angle MPO = angle KRO = angle P / 2 = 60/2 = 30 degrees.
a) Consider the IOC triangle: KMO angle = 25 degrees, MCO angle = 35 degrees. By the theorem on the sum of the angles of a triangle:
KMO angle + MKO angle + MOC angle = 180 degrees;
25 + 35 + IOC angle = 180;
angle IOC = 180 – 60;
angle MOK = 120 degrees.
b) Consider a ROC triangle: RKO angle = 35 degrees, KPO angle = 30 degrees. By the theorem on the sum of the angles of a triangle:
RKO angle + KRO angle + ROK angle = 180 degrees;
35 + 30 + ROC angle = 180;
angle ROC = 180 – 65;
ROCK angle = 115 degrees.
c) Consider a triangle MPE: angle RMO = 25 degrees, angle MPO = 30 degrees. By the theorem on the sum of the angles of a triangle:
angle RMO + angle MPO + angle MPO = 180 degrees;
25 + 30 + MOP angle = 180;
angle MOR = 180 – 55;
MOP angle = 125 degrees.
Answer: MOC angle = 120 degrees, ROK angle = 115 degrees, MOP angle = 125 degrees.