I have a feeling that thare is an easier or better way to do this but I couldn't think of it...
Each face of the tetrahedron is an equilateral triangle.
AM and BM are both heights of these equilateral triangles.
If we let the side of each equilateral triangle be x , then...
AB = x and AM = \(\frac{\sqrt3x}{2}\) and BM = \(\frac{\sqrt3x}{2}\)
And by the law of cosines....
\(\cos (\angle AMB)\,=\,\dfrac{-AB^2+AM^2+BM^2}{2(AM)(BM)}\\~\\ \cos (\angle AMB)\,=\,\dfrac{-x^2+(\frac{\sqrt3x}2)^2+(\frac{\sqrt3x}2)^2}{2(\frac{\sqrt3x}2)(\frac{\sqrt3x}2)}\\~\\ \cos (\angle AMB)\,=\,\dfrac{-x^2+\frac{3x^2}4+\frac{3x^2}4}{2(\frac{3x^2}4)}\\~\\ \cos (\angle AMB)\,=\,\dfrac{-1+\frac{3}4+\frac{3}4}{2(\frac{3}4)}\\~\\ \cos (\angle AMB)\,=\,\dfrac{-\frac44+\frac{3}4+\frac{3}4}{\frac{6}4}\\~\\ \cos (\angle AMB)\,=\,\frac{-4+3+3}{6}\\~\\ \cos (\angle AMB)\,=\,\frac13\)
By the Pythagorean identity...
\(\cos^2(\angle AMB)+\sin^2(\angle AMB)\,=\,1\\~\\ \frac{\cos^2(\angle AMB)}{\cos^2(\angle AMB)}+\frac{\sin^2(\angle AMB)}{\cos^2(\angle AMB)}\,=\,\frac{1}{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{(\frac13)^2}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\frac19}\\~\\ 1+\tan^2(\angle AMB)\,=\,9\\~\\ \tan^2(\angle AMB)\,=\,8\\~\\ \tan(\angle AMB)\,=\,\sqrt8\\~\\ \tan(\angle AMB)\,=\,2\sqrt2\)
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