In triangle ABC, angle C is 90 °, cosA = 0.48. Find sinB.

Since the angle C = 90 °, the triangle is right-angled, we denote its hypotenuse through c, then its legs will be equal:

c * cos (A) and c * sin (B).

By the Pythagorean theorem we get:

sin ^ 2 (A) + cos ^ 2 (B) = 1;

sin ^ 2 (B) = 1 – cos2 ^ (A) = 1 – (48/100) ^ 2 = 10000/10000 – 2304/1000 = 7696/10000.

sin (B) = 0.88.

Answer: The sought sine of angle B is 0.88.