In a right-angled triangle AC = 50, and the height of CH lowered to the hypotenuse is 8√32, find the sine of the angle b.

To solve the problem, let’s draw a triangle ABC, in which the angle C is a straight line:

Let’s draw the CH height to the hypotenuse. It is equal to 8√32.
The cosine of angle A is equal to the sine of angle B.
Consider the triangle AНС, the cosine of angle A is: √ (1 – sin2A) = √ (1 – sin28√32 / 50), and this is equal to (√113) / 25 or 0.43.
Then the sine of angle B is (√113) / 25.