In triangle ABC, angle C is 90 °, CH is the height, BC = 5, BH = 1. Find sin A.

Since CH is the height of the triangle ABC, the triangles BCH and AСН are rectangular.

In a right-angled triangle, the cosine of its acute angle is the ratio of the length of the adjacent leg to the length of the hypotenuse.

Then, in a right-angled triangle BCH:

CosСВН = ВН / ВC = ½, then CosABC = 1/5.

In a right-angled triangle, the cosine of one acute angle is equal to the sine of another acute angle, then SinBAC = CosABC = 1/5.

Answer: The sine of the ВCA angle is 1/5.