In a right-angled triangle ABC, angle B 90, AB 8cm, AC 16cm, find the angles that form the height

In a right-angled triangle ABC, angle B 90, AB 8cm, AC 16cm, find the angles that form the height of the BH with the legs of the triangle.

Consider the triangle ABC, AC – hypotenuse, AB and BC – legs, <B = 90º, AB = 8, AC = 16, BH – height drawn to the hypotenuse from angle B, <B = <ABH + <CBH. The height divides the triangle into right-angled triangles ABH and BHC.
Let’s define the angles of the triangle ABC:
By definition of the cosine of the angle of a right-angled triangle:
cos A = AB / AC = 8/16 = 1/2.
cos A = 1/2 corresponds to an angle of 60º.
<C = 180- <B- <A = 180-90-60 = 30º.
Consider the triangle ABH, <AHB = 90º, <BAH = 60º, AH – hypotenuse, AB and BH – legs, AB = 8.
From the sum of the angles of the triangle we have:
<ABH = 180- <AHB- <BAH = 180º-90º-60º = 30º.
Let’s go back to the triangle ABC, <B = <ABH + <CBH, find <CBH:
<CBH = <B- <ABH = 90-30 = 60º.
Answer: the height forms angles of 60º and 30º with the legs.