In triangle ABC, we have angle A = 90 degrees and sin B = (4/7). Find cos C. (No diagram was given)
as we know a triangle has 180 degrees we know that angles B and C must sum to 90.
so we have
\(\cos(C) = \cos(90^\circ - B) = \\ \cos(90^\circ)\cos(B) + \sin(90^\circ)\sin(B) = \\ \sin(B) = \dfrac 4 7\)
