# 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:

$a = 1 \ \\ h_{ 1 } = \sqrt{ a^2-(a/2)^2 } = \sqrt{ 1^2-(1/2)^2 } \doteq 0.866 \ \\ x = \dfrac{ 2 }{ 3 } \cdot \ h_{ 1 } = \dfrac{ 2 }{ 3 } \cdot \ 0.866 \doteq 0.5774 \ \\ \cos (A) = x/a \ \\ A = \dfrac{ 180^\circ }{ \pi } \cdot \arccos(x/a) = \dfrac{ 180^\circ }{ \pi } \cdot \arccos(0.5774/1) \doteq 54.7356 = 54.736 ^\circ = 54^\circ 44'8"$

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