1. The problem is to verify the six trigonometric ratios (sin θ, cos θ, tan θ, csc θ, sec θ, cot θ) for two right triangles given the side lengths. ### First Triangle (baseball ball trajectory): 2. Given sides: opposite = 21 m, adjacent = 28 m, hypotenuse = 35 m. 3. Calculate sin θ: $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{21}{35} = \frac{3}{5}$$ 4. Calculate cos θ: $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{28}{35} = \frac{4}{5}$$ 5. Calculate tan θ: $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{21}{28} = \frac{3}{4}$$ 6. Calculate csc θ (reciprocal of sin θ): $$\csc \theta = \frac{1}{\sin \theta} = \frac{35}{21} = \frac{5}{3}$$ 7. Calculate sec θ (reciprocal of cos θ): $$\sec \theta = \frac{1}{\cos \theta} = \frac{35}{28} = \frac{5}{4}$$ 8. Calculate cot θ (reciprocal of tan θ): $$\cot \theta = \frac{1}{\tan \theta} = \frac{28}{21} = \frac{4}{3}$$ Each matches the fractions given: sin 21/35, cos 28/35, tan 21/28, csc 35/21, sec 35/28, cot 28/21. ### Second Triangle (Bad Bunny's triangle): 9. Given sides: opposite = 18 m, hypotenuse = 30 m, find adjacent using Pythagoras: $$\text{adjacent} = \sqrt{30^2 - 18^2} = \sqrt{900 - 324} = \sqrt{576} = 24$$ 10. Calculate sin θ: $$\sin \theta = \frac{18}{30} = \frac{3}{5}$$ 11. Calculate cos θ: $$\cos \theta = \frac{24}{30} = \frac{4}{5}$$ 12. Calculate tan θ: $$\tan \theta = \frac{18}{24} = \frac{3}{4}$$ 13. Calculate csc θ: $$\csc \theta = \frac{1}{\sin \theta} = \frac{30}{18} = \frac{5}{3}$$ 14. Calculate sec θ: $$\sec \theta = \frac{1}{\cos \theta} = \frac{30}{24} = \frac{5}{4}$$ 15. Calculate cot θ: $$\cot \theta = \frac{1}{\tan \theta} = \frac{24}{18} = \frac{4}{3}$$ Each corresponds to the fractions given exactly. **Conclusion:** All given trigonometric ratios are correct for their respective triangles.