+188 ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of CD, then what is tan AMB?
+129852 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
