In triangle ABC, angle C is 90 °, sin A 4/5, AC = 9. Find AB

The sine of the angle in a right-angled triangle is equal to the ratio of the opposite leg to the hypotenuse, that is

sin A = BC / AB.

From here

BC = sin A * AB = 4 / 5AB.

By the Pythagorean theorem:

AB² = BC² + AC².

Knowing that BC = 4 / 5AB and AC = 9, we write this expression differently and solve the resulting equation:

AB² = (4 / 5AB) ² + 9²,

AB² = 16 / 25AB² + 81,

AB² – 16 / 25AB² = 81,

25 / 25AB² – 16 / 25AB² = 81,

9 / 25AB² = 81,

AB² = 81 / 9/25,

AB² = 81 * 25/9,

AB² = 225,

AB = ± √225,

AB = ± 15.

Since the length cannot be negative, AB = 15.

Answer: Side AB of triangle ABC is 15.