Question

prove that the smaller angle theta between any two diagonals of a cube is given by cos theta =1/3

Answer 1

Consider a cube of side, s.

The diagonals on each of the six faces = s√2.

The internal diagonal = [(s√2)^2 +s^2]^0.5. = [2s^2 + s^2[^0.5 = [3s^2]^0.5 = s√3.

The angle between the diagonal on the face and the internal diagonal is got as diagonal on each face/internal diagonal = arc cos √2/√3 = 35.26438968 deg.

The two internal diagonals intersect at 2x35.26438968 = 70.52877937 deg and cos 70.52877937 = 0.333333 or 1/3. Proved.

Answer 2

Step-by-step explanation:

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