The angle between line segment AH and the surface EFGH is approximately 24.2°.
The angle ([tex]\theta[/tex]), in sexagesimal degrees, lies between the line segment AH, in centimeters, and the line segment HF, in centimeters, whose length is found by the following trigonometric relation:
[tex]\theta = \cos^{-1} \left(\frac{HF}{HA} \right)[/tex] (1)
Where each length is determined by the following formulas:
[tex]HF = \sqrt{HG^{2}+FG^{2}}[/tex] (3)
[tex]HA = \sqrt{HF^{2}+FA^{2}}[/tex] (4)
If we know that [tex]HG = 17\,cm[/tex], [tex]FG = 5\,cm[/tex] and [tex]FA = 8\,cm[/tex], then the angle between AH and the surface EFGH is:
[tex]HF = \sqrt{(17\,cm)^{2}+(5\,cm)^{2}}[/tex]
[tex]HF \approx 17.7\,cm[/tex]
[tex]HA = \sqrt{(17.7\,cm)^{2}+(8\,cm)^{2}}[/tex]
[tex]HA \approx 19.4\,cm[/tex]
[tex]\theta \approx \cos^{-1} \left(\frac{17.7\,cm}{19.4\,cm} \right)[/tex]
[tex]\theta \approx 24.2^{\circ}[/tex]
The angle between line segment AH and the surface EFGH is approximately 24.2°.
To learn more on solids, we kindly invite to check this verified question: https://brainly.com/question/9740924
