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Right Triangle Trigonometery

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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 $$\tan\angle AMB$$?

Apr 19, 2018

#1
+6 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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Apr 20, 2018

#1
+6 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 Apr 20, 2018
#2
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Nice first question, and welcome to the forum!

Apr 20, 2018
#3
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Wow! Thanks hectictar! You are awesome at this!

Apr 20, 2018
#4
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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).

Apr 20, 2018
edited by Guest  Apr 20, 2018
edited by Guest  Apr 20, 2018