In a right-angled triangle, the median drawn to the hypotenuse divides the right angle

In a right-angled triangle, the median drawn to the hypotenuse divides the right angle into two angles, one of which is 8 times smaller than the other. find the sharp corners of a given triangular.

Let the value of the angle CBM = X0, then by condition, the angle ABM = 8 * X0.

The sum of the angles ABM and CBM is 90.

Angle ABM + CBM = 8 * X + X = 90.

9 * X = 90.

X = CBM = 90/9 = 10.

Angle ABM = 90 – 10 = 80.

Around the right-angled triangle ABC, you can describe a circle, the diameter of which coincides with the hypotenuse AC, and the median BM with the radius. Then BM = AM = CM = R, and the triangles ABM and BCM are isosceles.

Then the angle BCA = CBM = 10, angle BAC = ABM = 80.

Answer: The acute angles of the triangle ABC are equal to 10 and 80.