1. **Problem statement:** Given right triangles with two sides, find $\sin$, $\cos$, and $\tan$ of angles $A$ and $B$ as common fractions. 2. **Recall:** In a right triangle, $c$ is the hypotenuse, $a$ and $b$ are perpendicular sides, $A$ is opposite $a$, and $B$ opposite $b$. 3. **Formulas:** - $\sin A = \frac{a}{c}$, $\cos A = \frac{b}{c}$, $\tan A = \frac{a}{b}$ - $\sin B = \frac{b}{c}$, $\cos B = \frac{a}{c}$, $\tan B = \frac{b}{a}$ 4. **(i) Given:** $c=41$, $a=9$. Find $b$ using Pythagoras: $$b = \sqrt{c^2 - a^2} = \sqrt{41^2 - 9^2} = \sqrt{1681 - 81} = \sqrt{1600} = 40$$ 5. Calculate ratios: - $\sin A = \frac{9}{41}$ - $\cos A = \frac{40}{41}$ - $\tan A = \frac{9}{40}$ - $\sin B = \frac{40}{41}$ - $\cos B = \frac{9}{41}$ - $\tan B = \frac{40}{9}$ 6. **(ii) Given:** $c=37$, $a=35$. Find $b$: $$b = \sqrt{37^2 - 35^2} = \sqrt{1369 - 1225} = \sqrt{144} = 12$$ 7. Calculate ratios: - $\sin A = \frac{35}{37}$ - $\cos A = \frac{12}{37}$ - $\tan A = \frac{35}{12}$ - $\sin B = \frac{12}{37}$ - $\cos B = \frac{35}{37}$ - $\tan B = \frac{12}{35}$ 8. **(iii) Given:** $a=24$, $b=7$. Find $c$: $$c = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$$ 9. Calculate ratios: - $\sin A = \frac{24}{25}$ - $\cos A = \frac{7}{25}$ - $\tan A = \frac{24}{7}$ - $\sin B = \frac{7}{25}$ - $\cos B = \frac{24}{25}$ - $\tan B = \frac{7}{24}$ **Final answers:** (i) $\sin A = \frac{9}{41}$, $\cos A = \frac{40}{41}$, $\tan A = \frac{9}{40}$; $\sin B = \frac{40}{41}$, $\cos B = \frac{9}{41}$, $\tan B = \frac{40}{9}$. (ii) $\sin A = \frac{35}{37}$, $\cos A = \frac{12}{37}$, $\tan A = \frac{35}{12}$; $\sin B = \frac{12}{37}$, $\cos B = \frac{35}{37}$, $\tan B = \frac{12}{35}$. (iii) $\sin A = \frac{24}{25}$, $\cos A = \frac{7}{25}$, $\tan A = \frac{24}{7}$; $\sin B = \frac{7}{25}$, $\cos B = \frac{24}{25}$, $\tan B = \frac{7}{24}$.