In triangle ABC, sin A = 4/5. Find cos B.
\(\phantom{\sin A : \sin B : \sin C}\)
\(\phantom{5:6:4}\)
Let's write the problem like this: In triangle ABC, with C=90o, sin A = 4/5. Find cos B.
By Pythagoras, that is a 3-4-5 right triangle, thus the third side is 3.
cosine equals adjacent over hypotenuse ——>> cos(B) = 4 / 5
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