In a triangle ABC, the angle is c 90 °, sinA is three-fifths. find cos B

From the graphical representation of the problem, we find that AB is the hypotenuse, AC and BC are legs.
Let’s look at what sine and cosine are.
The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse, in our problem: sin A = a / c = BC / AB = 3/5.
The cosine of an acute angle in a right triangle is the ratio of the adjacent leg to the hypotenuse, in our problem: cos B = a / c = BC / AB = 3/5.
Therefore, cos B = sin A = 3/5.

Answer: cos B = 3/5.