In a rectangular triangle, the angle between the bisectrix and the median drawn from the vertex to the right

In a rectangular triangle, the angle between the bisectrix and the median drawn from the vertex to the right angle is 13 degrees. Find the larger of the two acute angles of the triangle.

Since the median BM is drawn from a right angle, it is equal to half of the hypotenuse AC, then BM = AM = CM, and therefore the triangles ABM and BKС are isosceles.

Since BK is the bisector of a right angle, then the angles KBC = ABK = 45. By condition, the angle MBK between the bisector and the median is 13. Then the angle MBC = KBC + MBK = 45 + 13 = 58.

Since the triangle BKС is isosceles, BM = MC, therefore, the angles MCB = ACB = MBC = 58.

Answer: The larger acute angle is 58.