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Yes, there is an almost-purely-geometric solution for this one. Insert the tetrahedron to the corners of a cube.
Answer is D.By cosine rule to triang. of sides 1, 3/8 & 3/8, cos(mu)= 1- 4/3 = -1/3.
Sorry.Side of tetrahedron a=1 and radius r=sqrt(6)/4, then: a^2 = 2r^2 - 2r^2cos(mu) -> cos(mu) = -1/3.Triangle 1,sqrt(3/8),sqrt(3/8).
Yes, there is an almost-purely-geometric solution for this one. Insert the tetrahedron to the corners of a cube.
ReplyDeleteAnswer is D.
ReplyDeleteBy cosine rule to triang. of sides 1, 3/8 & 3/8, cos(mu)= 1- 4/3 = -1/3.
Sorry.
ReplyDeleteSide of tetrahedron a=1 and radius r=sqrt(6)/4, then: a^2 = 2r^2 - 2r^2cos(mu) -> cos(mu) = -1/3.
Triangle 1,sqrt(3/8),sqrt(3/8).