In triangle ABC the right angle is at B, AB = 24 cm and BC = 10 cm.

Step 1. (a) $AC = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\ \text{cm}$.

Step 2. (b) $\sin A = \dfrac{BC}{AC} = \dfrac{10}{26} = \dfrac{5}{13}$, and $\cos C = \dfrac{BC}{AC} = \dfrac{5}{13}$. They are equal, because the side opposite $A$ is the side adjacent to $C$.

Step 3. (c) $\tan A = \dfrac{10}{24}$ and $\tan C = \dfrac{24}{10}$, so their product is $1$.