1. **State the problem:** We are given a right triangle with vertices A, B, and C, where angle C is the right angle. Sides are AC = 3, BC = 6\sqrt{2}, and hypotenuse AB = 9. We need to find the trigonometric ratios \(\sin, \cos, \tan\) for angles A and B. 2. **Recall the definitions (SOH-CAH-TOA):** - \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\) - \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\) - \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\) 3. **Identify sides relative to each angle:** - For angle A: - Opposite side: BC = \(6\sqrt{2}\) - Adjacent side: AC = 3 - Hypotenuse: AB = 9 - For angle B: - Opposite side: AC = 3 - Adjacent side: BC = \(6\sqrt{2}\) - Hypotenuse: AB = 9 4. **Calculate ratios for angle A:** \[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6\sqrt{2}}{9} = \frac{\cancel{3} \times 2\sqrt{2}}{\cancel{3} \times 3} = \frac{2\sqrt{2}}{3} \] \[ \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{9} = \frac{\cancel{3}}{\cancel{3} \times 3} = \frac{1}{3} \] \[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{6\sqrt{2}}{3} = \frac{\cancel{3} \times 2\sqrt{2}}{\cancel{3}} = 2\sqrt{2} \] 5. **Calculate ratios for angle B:** \[ \sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{9} = \frac{\cancel{3}}{\cancel{3} \times 3} = \frac{1}{3} \] \[ \cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6\sqrt{2}}{9} = \frac{\cancel{3} \times 2\sqrt{2}}{\cancel{3} \times 3} = \frac{2\sqrt{2}}{3} \] \[ \tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{6\sqrt{2}} = \frac{3}{\cancel{3} \times 2\sqrt{2}} = \frac{1}{2\sqrt{2}} \] 6. **Simplify \(\tan B\) by rationalizing the denominator:** \[ \tan B = \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2 \times 2} = \frac{\sqrt{2}}{4} \] **Final answers:** - \(\sin A = \frac{2\sqrt{2}}{3}\) - \(\cos A = \frac{1}{3}\) - \(\tan A = 2\sqrt{2}\) - \(\sin B = \frac{1}{3}\) - \(\cos B = \frac{2\sqrt{2}}{3}\) - \(\tan B = \frac{\sqrt{2}}{4}\)