1. **Problem statement:** We have three right triangles with given sides and angles, and we need to find $\sin A$, $\cos A$, and $\tan A$ for each triangle. 2. **Recall definitions:** For a right triangle with angle $A$: - $\sin A = \frac{\text{opposite side}}{\text{hypotenuse}}$ - $\cos A = \frac{\text{adjacent side}}{\text{hypotenuse}}$ - $\tan A = \frac{\text{opposite side}}{\text{adjacent side}}$ 3. **Triangle 1:** Given opposite side $=5$, adjacent side $=12$, hypotenuse $=13$. - Calculate $\sin A = \frac{5}{13}$. - Calculate $\cos A = \frac{12}{13}$. - Calculate $\tan A = \frac{5}{12}$. 4. **Triangle 2:** Given opposite side to $B$ is 4, hypotenuse $=7$, adjacent $=\sqrt{33}$. Since angle $A$ is complementary to angle $B$, opposite side to $A$ is adjacent to $B$, so opposite side to $A$ is $\sqrt{33}$, adjacent side to $A$ is 4. - Calculate $\sin A = \frac{\sqrt{33}}{7}$. - Calculate $\cos A = \frac{4}{7}$. - Calculate $\tan A = \frac{\sqrt{33}}{4}$. 5. **Triangle 3:** Given opposite side to $B$ is 8, adjacent side $=7$, side $BC=\sqrt{15}$, hypotenuse not labeled. - Since $BC=\sqrt{15}$ is the side opposite angle $A$ (assuming $BC$ is opposite $A$), and adjacent side is 7, hypotenuse $= \sqrt{7^2 + (\sqrt{15})^2} = \sqrt{49 + 15} = \sqrt{64} = 8$. - Calculate $\sin A = \frac{\sqrt{15}}{8}$. - Calculate $\cos A = \frac{7}{8}$. - Calculate $\tan A = \frac{\sqrt{15}}{7}$. **Final answers:** Triangle 1: $$\sin A = \frac{5}{13}, \quad \cos A = \frac{12}{13}, \quad \tan A = \frac{5}{12}$$ Triangle 2: $$\sin A = \frac{\sqrt{33}}{7}, \quad \cos A = \frac{4}{7}, \quad \tan A = \frac{\sqrt{33}}{4}$$ Triangle 3: $$\sin A = \frac{\sqrt{15}}{8}, \quad \cos A = \frac{7}{8}, \quad \tan A = \frac{\sqrt{15}}{7}$$