1. **Problem statement:** We have two right triangles. For the first triangle ABC with right angle at C, sides BC=8, AC=6, and hypotenuse AB=10, find the six trigonometric ratios for angles A and B. 2. **Formulas:** For any angle \(\theta\) in a right triangle: - \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\) - \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\) - \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\) 3. **Triangle ABC:** - Hypotenuse \(AB = 10\) - Side opposite \(A\) is \(BC = 8\) - Side adjacent to \(A\) is \(AC = 6\) - Side opposite \(B\) is \(AC = 6\) - Side adjacent to \(B\) is \(BC = 8\) 4. **Calculate ratios for angle A:** \[\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{10} = \frac{\cancel{8}}{\cancel{10}} = \frac{4}{5} = 0.8\] \[\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{10} = \frac{\cancel{6}}{\cancel{10}} = \frac{3}{5} = 0.6\] \[\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{6} = \frac{\cancel{8}}{\cancel{6}} = \frac{4}{3} \approx 1.3\] 5. **Calculate ratios for angle B:** \[\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{10} = \frac{3}{5} = 0.6\] \[\cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{10} = \frac{4}{5} = 0.8\] \[\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{8} = \frac{3}{4} = 0.75\] 6. **Summary for triangle ABC:** - \(\sin A = 0.8\), \(\cos A = 0.6\), \(\tan A = 1.3\) - \(\sin B = 0.6\), \(\cos B = 0.8\), \(\tan B = 0.75\) --- Since the user asked to solve the first question only, we stop here.