1. **State the problem:** We have a right triangle with sides $x=9$, $y=6$, and hypotenuse $z$. We need to find $\cos(\theta)$, $\sin(\theta)$, and $\tan(\theta)$ as exact fractions. 2. **Formula and rules:** For a right triangle with angle $\theta$: - $\cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}}$ - $\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$ - $\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$ 3. **Find the hypotenuse $z$ using the Pythagorean theorem:** $$ z = \sqrt{x^2 + y^2} = \sqrt{9^2 + 6^2} = \sqrt{81 + 36} = \sqrt{117} $$ 4. **Simplify $\sqrt{117}$:** $$ \sqrt{117} = \sqrt{9 \times 13} = 3\sqrt{13} $$ 5. **Calculate $\cos(\theta)$:** Since $\theta$ is at the bottom-right corner, the adjacent side to $\theta$ is $x=9$ and the hypotenuse is $z=3\sqrt{13}$. $$ \cos(\theta) = \frac{9}{3\sqrt{13}} = \frac{\cancel{9}}{\cancel{3}\sqrt{13}} = \frac{3}{\sqrt{13}} $$ 6. **Rationalize the denominator for $\cos(\theta)$:** $$ \cos(\theta) = \frac{3}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{3\sqrt{13}}{13} $$ 7. **Calculate $\sin(\theta)$:** The opposite side to $\theta$ is $y=6$. $$ \sin(\theta) = \frac{6}{3\sqrt{13}} = \frac{\cancel{6}}{\cancel{3}\sqrt{13}} = \frac{2}{\sqrt{13}} $$ 8. **Rationalize the denominator for $\sin(\theta)$:** $$ \sin(\theta) = \frac{2}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{2\sqrt{13}}{13} $$ 9. **Calculate $\tan(\theta)$:** $$ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{9} = \frac{2}{3} $$ **Final answers:** - $\cos(\theta) = \frac{3\sqrt{13}}{13}$ - $\sin(\theta) = \frac{2\sqrt{13}}{13}$ - $\tan(\theta) = \frac{2}{3}$