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\(ABCD\) is a regular tetrahedron (right pyramid whose faces are all equilateral triangles). If \(M\) is the midpoint of \(\overline{CD}\), then what is \(\cos \angle ABM\)?

 May 14, 2020
 #1
avatar+111360 
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We can use  the Law of Cosines to solve this

 

Call the side length of the tetrahedron, S

 

The slant  height  =  height of the equilateral triangle  = AM  =  BM  =   (√3/2)S

AB = the side length

 

So  we have that

 

AM^2  =  BM^2  + AB^2  -  2 ( BM * AB) *  cos (ABM)

 

(3/4)S^2  =  (3/4)S^2  + S^2 - 2 ( 3/4)S^2  * cos (ABM)  

 

0  =  S^2 - (3/2)S^2 * cos (ABM)       divide through by S^2

 

0 = 1 - (3/2)cos (ABM) 

  

-1 = -(3/2)cos(ABM) 

 

(2/3)  = cos (ABM)

 

 

cool cool cool

 May 14, 2020
 #2
avatar+67 
0

thx really.

 

I really couldn't graph it out! 

 

Thanks,

 

-Zach

 May 15, 2020

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