Consider a square-based straight pyramid. Suppose that the base is a square with sides 10 km long, and all other edges are 17 km long. Find an approximate value of the angle formed between the base and a triangular face. Present your answer in degrees, accurate up to four or more decimal places.
Answers
Answer:
72.06
Step-by-step explanation:
Given that :
Side Lenght, a = 10
Edge length, b = 17
Using Pythagoras :
h = height of pyramid
h = sqrt (b² - 1/2a²)
h = sqrt(17^2 - 10^2/2)
h = sqrt(289 - 50)
h = 15.46
Slant height :
L = sqrt(h² + a²/4)
L = sqrt(15.46² + 25)
L = sqrt(239 + 25)
L = sqrt(264)
L = 16.25
Sinα = h/L
α = sin^-1 (15.46/16.25)
α = 72.060933
= 72.06
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°.
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