In triangle $ABC,$ $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}.$  Find $\cos C.$
Question:
In triangle $ABC,$ $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}.$  Find $\cos C.$
Answer:
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\frac{16}{65}
In triangle $ABC,$ $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}.$  Find $\cos C.$
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