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ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of CD, then what is tan AMB?

Mathgenius  Nov 18, 2018
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Let the side of the tetrahedron   =   s

 

Since  each face is an equilateral triangle......the slant height, h is given by

 

sin (60°)  = h / s ⇒     h = * sin (60°)  =  √3/2 * s = √3/2 

 

And he distance from M to the center of the triangle is given by

 

tan (30°) = C / [ (1/2)s ] ⇒  C = (1/2)s * (1/√3)  =  1 / [ 2√3]  =  √3/6 *s   (1)

 

And....using the Pythagorean Theorem, we can find the height of the triangle as

 

√ [  ( √3/2* s)^2  - (√3/6*s)^2 ]  =

 

s√ [ 3/4  - 3/36 ]  =

 

s√ [ 27 - 3] /36] =

 

s√[24/36] =

 

s√[2/3]     (2)

 

So....the tangent of AMB  =  (2) / (1)  =  s√[2/3]  / [ √3/6 *s] =

 

√2              6               6√2          

__    x      ___   =      _____  =       2√2     

√3             √3               3 

 

 

cool cool cool    

CPhill  Nov 18, 2018

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