1. **Problem 1:** Given a right triangle with angle $\theta$, opposite side = 9, adjacent side = 40. 2. **Find:** $\sin \theta$, $\csc \theta$, $\cos \theta$, $\sec \theta$, $\tan \theta$, $\cot \theta$. 3. **Step 1:** Calculate the hypotenuse $h$ using Pythagoras theorem: $$h = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41$$ 4. **Step 2:** Use definitions of trig functions: - $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{41}$ - $\csc \theta = \frac{1}{\sin \theta} = \frac{41}{9}$ - $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{40}{41}$ - $\sec \theta = \frac{1}{\cos \theta} = \frac{41}{40}$ - $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{9}{40}$ - $\cot \theta = \frac{1}{\tan \theta} = \frac{40}{9}$ --- 5. **Problem 2:** Given a right triangle with angle $\theta$, opposite side = 7, hypotenuse = 11. 6. **Find:** $\sin \theta$, $\csc \theta$, $\cos \theta$, $\sec \theta$, $\tan \theta$, $\cot \theta$. 7. **Step 1:** Calculate the adjacent side $a$ using Pythagoras theorem: $$a = \sqrt{11^2 - 7^2} = \sqrt{121 - 49} = \sqrt{72} = 6\sqrt{2}$$ 8. **Step 2:** Use definitions of trig functions: - $\sin \theta = \frac{7}{11}$ - $\csc \theta = \frac{11}{7}$ - $\cos \theta = \frac{6\sqrt{2}}{11}$ - $\sec \theta = \frac{11}{6\sqrt{2}}$ - $\tan \theta = \frac{7}{6\sqrt{2}}$ - $\cot \theta = \frac{6\sqrt{2}}{7}$ **Summary:** Problem 1: $\sin \theta = \frac{9}{41}$, $\csc \theta = \frac{41}{9}$, $\cos \theta = \frac{40}{41}$, $\sec \theta = \frac{41}{40}$, $\tan \theta = \frac{9}{40}$, $\cot \theta = \frac{40}{9}$ Problem 2: $\sin \theta = \frac{7}{11}$, $\csc \theta = \frac{11}{7}$, $\cos \theta = \frac{6\sqrt{2}}{11}$, $\sec \theta = \frac{11}{6\sqrt{2}}$, $\tan \theta = \frac{7}{6\sqrt{2}}$, $\cot \theta = \frac{6\sqrt{2}}{7}$