In triangle ABC it is known that angle A = 44 degrees, angle B = 56 degrees.

In triangle ABC it is known that angle A = 44 degrees, angle B = 56 degrees. The bisectors AK and BM of the triangle meet at point O. Find the corners of the quadrilateral AOBC.

1. In △ ABC:
∠A + ∠B + ∠C = 180 ° (by the theorem on the sum of the angles of a triangle);
44 ° + 56 ° + ∠C = 180 °;
∠C = 180 ° – 100 °;
∠C = 80 °.
In a quadrilateral AOBC ∠ACB = ∠C = 80 °.
2. Since AK is the bisector of ∠A, then AK divides ∠A in half:
∠OAB = ∠OAC = ∠A / 2 = 44 ° / 2 = 22 °.
3. Since BM is the bisector of ∠B, BM divides ∠B in half:
∠OBA = ∠OBC = ∠B / 2 = 56 ° / 2 = 28 °.
4. In the quadrilateral AOBC by the theorem on the sum of the angles of the quadrilateral:
∠OAC + ∠AOB + ∠OBC + ∠ACB = 360 °;
22 ° + ∠AOB + 28 ° + 80 ° = 360 °;
∠AOB = 360 ° – 130 °;
∠AOB = 230 °.
Answer: ∠OAC = 22 °, ∠AOB = 230 °, ∠OBC = 28 °, ∠ACB = 80 °.