The points of tangency of the sides of a triangle with a circle inscribed in it divide the circle into arcs

The points of tangency of the sides of a triangle with a circle inscribed in it divide the circle into arcs in the ratio 10:11:15. Find the corners of this triangle.

Given △ ABC, O is the center of the inscribed circle, M, N and K are the tangency points of the circle with sides AB, BC and AC, respectively.
1. Arc MN + Arc NK + Arc MK = 360 °.
Let x be the coefficient of proportionality, then:
10 * x + 11 * x + 15 * x = 360 °;
36 * x = 360 °;
x = 360 ° / 36;
x = 10 °.
Thus:
Arc MN = 10 * x = 10 * 10 ° = 100 °;
Arc NK = 11 * x = 11 * 10 ° = 110 °;
Arc MK = 15 * x = 15 * 10 ° = 150 °.
2. Draw radii from point O to the points of tangency.
2.1. Consider AMOK: ∠AMO = ∠AKO = 90 ° (since the radius drawn to the tangent point from the center of the circle is perpendicular to the tangent), ∠MOK = 150 ° (since the degree measure of the central angle is equal to the degree measure of the arc on which it rests) …
By the theorem on the sum of the angles of a quadrangle:
∠KAM + ∠AMO + ∠MOK + ∠AKO = 360 °;
∠KAM + 90 ° + 150 ° + 90 ° = 360 °;
∠KAM + 330 ° = 360 °;
∠KAM = 360 ° – 330 °;
∠KAM = 30 °.
∠KAM = ∠A = 30 °.
2.2. Consider BMON: ∠BMO = ∠BNO = 90 °, ∠MON = 100 °.
By the theorem on the sum of the angles of a quadrangle:
∠MBN + ∠BMO + ∠MON + ∠BNO = 360 °;
∠MBN + 90 ° + 100 ° + 90 ° = 360 °;
∠MBN + 280 ° = 360 °;
∠MBN = 360 ° – 280 °;
∠MBN = 80 °.
∠MBN = ∠B = 80 °.
2.3. Consider CNOK: ∠CKO = ∠CNO = 90 °, ∠KON = 110 °.
By the theorem on the sum of the angles of a quadrangle:
∠NCK + ∠CKO + ∠KON + ∠CNO = 360 °;
∠NCK + 90 ° + 110 ° + 90 ° = 360 °;
∠NCK + 290 ° = 360 °;
∠NCK = 360 ° – 290 °;
∠NCK = 70 °.
∠NCK = ∠C = 70 °.
Answer: ∠A = 30 °, ∠B = 80 °, ∠C = 70 °.