In the triangle ABC, the angle is C = 90 °, CH-height, BC = 10, cos A = √24 / 5. Find AH.

Let’s write down the basic trigonometric identity and find the sine of the angle A.
sin² A + cos² A = 1
sin A = √ (1 – cos² A) = √ (1 – 24/25) = √1 / 25 = 1/5.

The sine of angle A is the ratio of the opposite leg BC to the hypotenuse AB. Find the hypotenuse AB:
AB = BC / sin A = 10 / 1/5 = 10 * 5 = 50.
Using the property of a leg in a right-angled triangle, we find the projection of this leg:
BC² = BH * AB
100 = BH * 50
BH = 2
Find AH:
AH = AB – BH = 50 – 2 = 48.
Answer: AH = 48.