Tetrahedron

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

Correct answer:

A =  54.7356 °

Step-by-step explanation:

$$\begin{gathered} a=1 \\ {h}_1=\sqrt{a^2-(a\mathrm{/}2{)}^2}=\sqrt{1^2-(1\mathrm{/}2{)}^2}\doteq 0.866 \\ x={\displaystyle \frac{2}{3}}\cdot h_1={\displaystyle \frac{2}{3}}\cdot 0.866\doteq 0.5774 \\ \cos (A)=x\mathrm{/}a \\ A={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arccos (x\mathrm{/}a)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arccos (0.5774\mathrm{/}1)=54.7356^{\circ }=54^{\circ }44^{\mathrm{\prime }}8\mathrm{"} \end{gathered}$$

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