+177 \(ABCD\) is a regular tetrahedron (right pyramid whose faces are all equilateral triangles). If \(M\) is the midpoint of \(\overline{CD}\), then what is \(\cos \angle ABM\)?
+129852 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)
