1. **Problem:** A road increases 8 m in altitude for every 100 m of horizontal distance. Calculate the angle of inclination of the road, to the nearest tenth of a degree. 2. **Formula:** The angle of inclination $\theta$ can be found using the tangent function: $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$ where opposite is the vertical rise and adjacent is the horizontal distance. 3. **Calculation:** $$\tan \theta = \frac{8}{100}$$ 4. To find $\theta$, take the arctangent (inverse tangent): $$\theta = \tan^{-1}\left(\frac{8}{100}\right)$$ 5. Calculate the value: $$\theta = \tan^{-1}(0.08) \approx 4.57^\circ$$ 6. **Answer:** Rounded to the nearest tenth, $$\theta \approx 4.6^\circ$$ --- 7. **Problem:** For $\angle M$ in the triangle with sides $LK=16$ (vertical), $KM=65$ (hypotenuse), and right angle at $L$: a) Label the hypotenuse, opposite, and adjacent sides relative to $\angle M$. b) Determine $\tan M$ to the nearest hundredth. 8. **Step a:** - Hypotenuse: side opposite the right angle, longest side $KM=65$. - Opposite side to $\angle M$: side opposite $M$ is $LK=16$ (vertical side). - Adjacent side to $\angle M$: the side next to $M$ that is not the hypotenuse, which is $LM$ (unknown length). 9. **Step b:** Calculate $LM$ using Pythagoras theorem: $$LM = \sqrt{KM^2 - LK^2} = \sqrt{65^2 - 16^2} = \sqrt{4225 - 256} = \sqrt{3969} = 63$$ 10. Calculate $\tan M$: $$\tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{LK}{LM} = \frac{16}{63}$$ 11. Simplify fraction with cancellation: $$\tan M = \frac{\cancel{16}}{\cancel{63}}$$ (no common factors to cancel) 12. Calculate decimal value: $$\tan M \approx 0.25397 \approx 0.25$$ 13. **Answer:** - Hypotenuse: $KM=65$ - Opposite side to $\angle M$: $LK=16$ - Adjacent side to $\angle M$: $LM=63$ - $\tan M \approx 0.25$