1. **State the problem:** Find the indicated trigonometric ratios as fractions for the given right triangles XYZ and ABC. 2. **Recall the SOH CAH TOA rules:** - $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ 3. **Triangle XYZ:** Right angle at Y, sides given: - $XY = 32$ (adjacent to angle X) - $YZ = 24$ (opposite to angle X) Calculate hypotenuse $XZ$ using Pythagoras: $$XZ = \sqrt{XY^2 + YZ^2} = \sqrt{32^2 + 24^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40$$ 4. **Triangle ABC:** Right angle at B, sides given: - $AB = 18$ (adjacent to angle A) - $BC = 30$ (hypotenuse) Calculate opposite side $AC$ using Pythagoras: $$AC = \sqrt{BC^2 - AB^2} = \sqrt{30^2 - 18^2} = \sqrt{900 - 324} = \sqrt{576} = 24$$ 5. **Find each ratio:** 1. $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{24}{30}$ 2. $\tan C$ in triangle ABC: angle C is complementary to angle A, so $\tan C = \frac{\text{opposite to C}}{\text{adjacent to C}} = \frac{AB}{AC} = \frac{18}{24}$ 3. $\cos X = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{XZ} = \frac{32}{40}$ 4. $\sin Z$ in triangle XYZ: angle Z is complementary to angle X, so $\sin Z = \cos X = \frac{32}{40}$ 5. $\tan Z = \frac{\text{opposite}}{\text{adjacent}} = \frac{XY}{YZ} = \frac{32}{24}$ 6. $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{18}{30}$ 7. $\sin X = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{YZ}{XZ} = \frac{24}{40}$ 8. $\cos C = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{24}{30}$ 9. $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{AB} = \frac{24}{18}$ 10. $\tan X = \frac{\text{opposite}}{\text{adjacent}} = \frac{YZ}{XY} = \frac{24}{32}$ 6. **Simplify fractions with cancellation:** - $\sin A = \frac{24}{30} = \frac{\cancel{6} \times 4}{\cancel{6} \times 5} = \frac{4}{5}$ - $\tan C = \frac{18}{24} = \frac{\cancel{6} \times 3}{\cancel{6} \times 4} = \frac{3}{4}$ - $\cos X = \frac{32}{40} = \frac{\cancel{8} \times 4}{\cancel{8} \times 5} = \frac{4}{5}$ - $\sin Z = \frac{32}{40} = \frac{4}{5}$ (same as $\cos X$) - $\tan Z = \frac{32}{24} = \frac{\cancel{8} \times 4}{\cancel{8} \times 3} = \frac{4}{3}$ - $\cos A = \frac{18}{30} = \frac{\cancel{6} \times 3}{\cancel{6} \times 5} = \frac{3}{5}$ - $\sin X = \frac{24}{40} = \frac{\cancel{8} \times 3}{\cancel{8} \times 5} = \frac{3}{5}$ - $\cos C = \frac{24}{30} = \frac{4}{5}$ (same as $\sin A$) - $\tan A = \frac{24}{18} = \frac{\cancel{6} \times 4}{\cancel{6} \times 3} = \frac{4}{3}$ - $\tan X = \frac{24}{32} = \frac{3}{4}$ **Final answers:** 1. $\sin A = \frac{4}{5}$ 2. $\tan C = \frac{3}{4}$ 3. $\cos X = \frac{4}{5}$ 4. $\sin Z = \frac{4}{5}$ 5. $\tan Z = \frac{4}{3}$ 6. $\cos A = \frac{3}{5}$ 7. $\sin X = \frac{3}{5}$ 8. $\cos C = \frac{4}{5}$ 9. $\tan A = \frac{4}{3}$ 10. $\tan X = \frac{3}{4}$