Tetrahedron

What is the angle of the sides from the base of a three-sided pyramid where the sides are identical?

Result

A =  54.736 °

Solution:

a=1 h1=a2−(a/2)2=12−(1/2)2≐0.866 x=23⋅ h1=23⋅ 0.866≐0.5774 cos⁡(A)=x/a A=180∘π⋅arccos⁡(x/a)=180∘π⋅arccos⁡(0.5774/1)≐54.7356=54.736∘=54∘44′8"a = 1 \ \\ h_{ 1 } = \sqrt{ a^2-(a/2)^2 } = \sqrt{ 1^2-(1/2)^2 } \doteq 0.866 \ \\ x = \dfrac{ 2 }{ 3 } \cdot \ h_{ 1 } = \dfrac{ 2 }{ 3 } \cdot \ 0.866 \doteq 0.5774 \ \\ \cos (A) = x/a \ \\ A = \dfrac{ 180^\circ }{ \pi } \cdot \arccos(x/a) = \dfrac{ 180^\circ }{ \pi } \cdot \arccos(0.5774/1) \doteq 54.7356 = 54.736 ^\circ = 54^\circ 44'8"

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