In triangle ABC, BC = 20 cm, AC = 16 cm, sinA = 3/8, find sinB.

To find the sine of angle B, we use the theorem of sines, according to which in any triangle the ratios of the length of each side to the sine of the opposite angle are equal to each other.

In the initial data for this task, it is reported that in this triangle ABC, the length of the BC side is 20 cm, the AC side length is 16 cm, and the sine of the angle A opposite to the BC side is 3/8, therefore, applying the theorem of sines, we can write the following relation :

| BC | / sin (A) = | AC | / sin (B),

whence follows:

sin (B) = | AC | * sin (A) / | BC | = 16 * (3/8) / 20 = 6/20 = 3/10 = 0.3.

Answer: sin (B) = 0.3.