In a right-angled triangle, the angle between the height and the bisector drawn from the vertex of the right

In a right-angled triangle, the angle between the height and the bisector drawn from the vertex of the right angle is 40 degrees. Find the larger angle of the given triangle.

The СК bisector divides the right angle ACB in half:
<ACK = <KCB = 90 ° / 2 = 45 °.
<MCK = 40 ° (angle between the SC bisector and CM height).
The height of the CM forms a right angle BMC with the hypotenuse AB.
In a right-angled triangle MBC <MCB = <KCB + <MCK = 45 ° + 40 ° = 85 °.
<MBC = 180 ° – <BMC – <MCB = 180 ° – 90 ° – 85 ° = 5 °.
<MBC = <ABC = 5 °.
In a right-angled triangle ABC <BCA = 180 ° – <ACB – <ABC = 180 ° – 90 ° – 5 ° = 85 °.
Answer: The larger acute angle of the triangle is 85 °.