The approximate value of the angle formed between the base and a triangular face is 72.0775°.
Let [tex]h[/tex] be the distance between the center of the square base and the midpoint of the side of the square and let [tex]H[/tex] the height of the triangular face of the pyramid. By properties of the square and the Pythagorean theorem we derive expressions for [tex]h[/tex] and [tex]H[/tex], respectively:
Square base
[tex]h = 0.5\cdot L[/tex] (1)
Triangular face
[tex]H = \sqrt{l^{2}-0.25\cdot L^{2}}[/tex] (2)
Where:
- [tex]L[/tex] - Side length of the square base, in kilometers.
- [tex]l[/tex] - Remaining side length of the triangular face, in kilometers.
Finally, by trigonometry we can derive an expression for the angle between the base and a triangular face ([tex]\theta[/tex]), in sexagesimal degrees:
[tex]\theta = \cos^{-1} \frac{h}{H}[/tex] (3)
If we know that [tex]L = 10\,km[/tex] and [tex]l = 17\,km[/tex], then the angle between the base and a triangular face is:
[tex]h = 5\,km[/tex]
[tex]H = \sqrt{17^{2}-5^{2}}[/tex]
[tex]H \approx 16.248\,km[/tex]
[tex]\theta = \cos^{-1} \frac{5\,km}{16.248\,km}[/tex]
[tex]\theta \approx 72.0775^{\circ}[/tex]
The approximate value of the angle formed between the base and a triangular face is 72.0775°.
We kindly invite to check this question on pyramids: https://brainly.com/question/16303112
