In triangle $ABC$, we have $\angle A = 90^\circ$ and \(\sin B = \frac{4}{9}\). Find $\cos C$.
Find cos C.
Hello Guest!
If the sin B = 4/9 in a right triangle, then the ratio of the sides is
a: b: c = 9: 4: 3.
So is
cos C = 4/9
!
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