The edge length of a regular tetrahedron ABCD is 2. M and N are the midpoints of BC and AD, respectively. Find angle AMD.
+130071 Point N is irrelevant here
Since triangle ABC is equliateral with sides of 2, its height = the slant height of the tetrahedron = sqrt (3) = AM
Call the center of the base E
The distance across the base also = sqrt 3 .....so 1/2 of this distance = sqrt (3) / 2
And AME forms a right triangle with leg ME = sqrt (3) / 2 and hypotenuse AM = sqrt (3)
Then cos AME = cos AMD = ME / AM = [sqrt (3) / 2 ] /sqrt (3) = 1/2
So
arccos (1/2) = measure of AMD = 60°
