One of the corners of a right-angled triangle is equal to 47, find the angle between the hypotenuse

One of the corners of a right-angled triangle is equal to 47, find the angle between the hypotenuse and the median drawn from the vertex of the right angle.

ABC – right-angled triangle, angle C = 90 degrees, angle CAB = 47 degrees, CM – median.
1. From the properties of a right-angled triangle, it is known that the median drawn from a right angle to the hypotenuse is equal to half of the hypotenuse. Then:
CM = AB / 2.
Also, the median CM divides the hypotenuse AB into 2 equal segments:
AB = MB = AB / 2.
Then:
CM = AM = MB.
2. Consider a triangle AMC: AM = CM, then AMC is an isosceles triangle with sides AM and CM and the base of the AC. The angles CAM (angle CAB) and MCA are the angles pi at the base of an isosceles triangle, hence the angle CAM = angle MCA = 47 degrees.
By the theorem on the sum of the angles of a triangle:
CAM angle + AMC angle + MCA angle = 180 degrees;
47 + angle AMC + 47 = 180;
angle AMC = 180 – 94;
angle AMC = 86 degrees.
3. The angles of the AMC and BMC are adjacent, therefore:
AMC angle + BMC angle = 180 degrees;
86 + BMC angle = 180;
BMC angle = 180 – 86;
BMC angle = 94 degrees.
Answer: AMC angle = 56 degrees, BMC angle = 94 degrees.