$ABCD$ is a regular tetrahedron (right pyramid whose faces are all equilateral triangles).  If $M$ is the midpoint of $\overline{CD}$, then what is $\cos \angle ABM$?
Question:
$ABCD$ is a regular tetrahedron (right pyramid whose faces are all equilateral triangles).  If $M$ is the midpoint of $\overline{CD}$, then what is $\cos \angle ABM$?
Answer:
Let's think step by step.

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\frac{\sqrt{3}}{3}
$ABCD$ is a regular tetrahedron (right pyramid whose faces are all equilateral triangles).  If $M$ is the midpoint of $\overline{CD}$, then what is $\cos \angle ABM$?
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