1. **State the problem:** Given a right triangle with an adjacent leg of length 7 cm and a hypotenuse of length 13 cm, find all six trigonometric ratios for angle $\theta$. 2. **Find the opposite leg:** Use the Pythagorean theorem $$\text{opposite} = \sqrt{\text{hypotenuse}^2 - \text{adjacent}^2} = \sqrt{13^2 - 7^2} = \sqrt{169 - 49} = \sqrt{120} = 2\sqrt{30}.$$ 3. **Calculate the six trigonometric ratios:** - $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2\sqrt{30}}{13}.$ - $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{13}.$ - $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2\sqrt{30}}{7}.$ - $\csc \theta = \frac{1}{\sin \theta} = \frac{13}{2\sqrt{30}} = \frac{13\sqrt{30}}{60}.$ - $\sec \theta = \frac{1}{\cos \theta} = \frac{13}{7}.$ - $\cot \theta = \frac{1}{\tan \theta} = \frac{7}{2\sqrt{30}} = \frac{7\sqrt{30}}{60}.$ 4. **Summary:** $$\sin \theta = \frac{2\sqrt{30}}{13}, \quad \cos \theta = \frac{7}{13}, \quad \tan \theta = \frac{2\sqrt{30}}{7},$$ $$\csc \theta = \frac{13\sqrt{30}}{60}, \quad \sec \theta = \frac{13}{7}, \quad \cot \theta = \frac{7\sqrt{30}}{60}.$$