1. **Stating the problem:** We are given a right triangle with vertical side AB of height $1.6 + 25 = 26.6$ meters, and angle $\angle ACB = 38^\circ$ between the diagonal AC and the horizontal BC. We want to find the length of AC. 2. **Formula used:** In a right triangle, the cosine of an angle is the ratio of the adjacent side over the hypotenuse: $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ Here, $\theta = 38^\circ$, adjacent side is BC, hypotenuse is AC. 3. **Important rules:** - The vertical side AB is opposite to angle $38^\circ$. - The Pythagorean theorem relates sides: $$AC^2 = AB^2 + BC^2$$ 4. **Find BC using tangent:** Since $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$, we have: $$\tan(38^\circ) = \frac{AB}{BC} \Rightarrow BC = \frac{AB}{\tan(38^\circ)}$$ 5. **Calculate BC:** $$BC = \frac{26.6}{\tan(38^\circ)}$$ Using $\tan(38^\circ) \approx 0.7813$: $$BC \approx \frac{26.6}{0.7813} \approx 34.06$$ 6. **Calculate AC using Pythagoras:** $$AC = \sqrt{AB^2 + BC^2} = \sqrt{26.6^2 + 34.06^2}$$ Calculate squares: $$26.6^2 = 707.56, \quad 34.06^2 = 1160.08$$ Sum: $$707.56 + 1160.08 = 1867.64$$ 7. **Final length of AC:** $$AC = \sqrt{1867.64} \approx 43.22$$ **Answer:** The length of the diagonal AC is approximately $43.22$ meters.