+295 Hello web2.0calc people! I was recommended here by a friend and looking at it, its pretty cool 😎!
My first question,
\(ABCD\) is a regular tetrahedron (right triangular pyramid). If \(M\) is the midpoint of \(\overline{CD}\), what is \(\tan\angle AMB\)?
+9466 
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...
\(\cos^2(\angle AMB)+\sin^2(\angle AMB)\,=\,1\\~\\ \frac{\cos^2(\angle AMB)}{\cos^2(\angle AMB)}+\frac{\sin^2(\angle AMB)}{\cos^2(\angle AMB)}\,=\,\frac{1}{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{(\frac13)^2}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\frac19}\\~\\ 1+\tan^2(\angle AMB)\,=\,9\\~\\ \tan^2(\angle AMB)\,=\,8\\~\\ \tan(\angle AMB)\,=\,\sqrt8\\~\\ \tan(\angle AMB)\,=\,2\sqrt2\)
+9466
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...
\(\cos^2(\angle AMB)+\sin^2(\angle AMB)\,=\,1\\~\\ \frac{\cos^2(\angle AMB)}{\cos^2(\angle AMB)}+\frac{\sin^2(\angle AMB)}{\cos^2(\angle AMB)}\,=\,\frac{1}{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\cos^2(\angle AMB)}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{(\frac13)^2}\\~\\ 1+\tan^2(\angle AMB)\,=\,\frac1{\frac19}\\~\\ 1+\tan^2(\angle AMB)\,=\,9\\~\\ \tan^2(\angle AMB)\,=\,8\\~\\ \tan(\angle AMB)\,=\,\sqrt8\\~\\ \tan(\angle AMB)\,=\,2\sqrt2\)
Hectictar:
Let the side length of each equilateral triangle = 1
Slant height of a regular tetrahedron =1 x [sqrt(3) /2]
Altitude(height) of regular tetrahedron =[1 x sqrt(6)] / 3 [From A to meet MB at N]
Draw atraight line From M to meet the height at N [Now you have a right triangle AMN]
This straight line MN =[sqrt(3) /2]^2 - [sqrt(6)/3]^2 =1/12 [By Pythagoras]
This straight line MN =sqrt(1/12)
Tan(AMN) =AMB =[sqrt(6) / 3] / [sqrt(1/12)] =sqrt(72)/3 =6sqrt(2)/3 =2sqrt(2).
