### Instruction: In triangle $ABC$, we have $\angle A = 90^\circ$ and $\sin B = \frac{4}{7}$.  Find $\cos C$.### Response: I know that in a right triangle, the sine of one acute angle is the cosine of the other acute angle, and vice versa.
So, if $\sin B = \frac{4}{7}$, then $\cos C = \frac{4}{7}$ as well.
I don't need to use any other information, such as the side lengths, since the problem only asks for the cosine of the angle.