Points M, N and K are selected on the circle so that the arc MN = 80 °, the arc MK = 140 °

Points M, N and K are selected on the circle so that the arc MN = 80 °, the arc MK = 140 °. Find the angles of a triangle ABC whose sides touch the circle at points M, N and K.

Let us denote by O the center of the circle. The radii OM, ON, OK intersect in it.
These radii form right angles with the sides of the ABC triangle.
At the intersection of OM and OK sides AB and AC, we have formed a quadrangle AMOK.
Let’s find its angle MAC:
∠MAK = 360 ° – 90 ° – 90 ° – 140 ° = 40 °.
At the intersection of ОМ and ОN sides AB and BC, we have formed a quadrangle MVNO.
Let’s find its angle MBN:
∠МВN = 360 ° – 90 ° – 90 ° – 80 ° = 100 °.
At the intersection of OK and ON sides of AC and BC, we have formed a quadrangle SKNO.
Let’s find its angle КСN:
∠KSN = 360 ° – 90 ° – 90 ° – (360 ° – 140 ° – 80 °) = 360 ° – 90 ° – 90 ° – 140 ° = 40 °.
ANSWER: ∠А = 40 °, ∠В = 100 °, ∠С = 40 °.