Right Triangle Abc 8Ef942

1. **State the problem:** We have a right triangle $ABC$ with right angle at $B$, vertical side $AB = 8$, horizontal side $BC = 6$, and hypotenuse $AC$. We want to find the length of the hypotenuse $AC$ and the angles $\alpha$ and $\theta$. 2. **Formula used:** In a right triangle, the Pythagorean theorem states: $$AC^2 = AB^2 + BC^2$$ Also, to find angles, we use trigonometric ratios: $$\sin \alpha = \frac{opposite}{hypotenuse}, \quad \cos \alpha = \frac{adjacent}{hypotenuse}, \quad \tan \alpha = \frac{opposite}{adjacent}$$ 3. **Calculate the hypotenuse $AC$:** $$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ 4. **Calculate angle $\alpha$ at $A$:** Using tangent: $$\tan \alpha = \frac{opposite}{adjacent} = \frac{AB}{BC} = \frac{8}{6} = \frac{4}{3}$$ So, $$\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$$ 5. **Calculate angle $\theta$ at $C$:** Since the triangle's angles sum to $180^\circ$ and $B$ is $90^\circ$: $$\theta = 90^\circ - \alpha = 90^\circ - 53.13^\circ = 36.87^\circ$$ **Final answers:** - Hypotenuse $AC = 10$ - Angle $\alpha \approx 53.13^\circ$ - Angle $\theta \approx 36.87^\circ$