# How to do this? :(

1) A line makes angles A,B,C and D with four diagonals of a cube (cosA)^2 + (cosB)^2 + (cosC)^2 + (cosD)^2 = ?

2)A line makes angles A,B,C and D with two diagonals of a cube (cosA)^2 + (cosB)^2 + (cosC)^2 + (cosD)^2 = ?

Note by Christian Daang
3 years, 8 months ago

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Can you check my solution also?

In #2, can you check if my solution is correct?

The Diagonals of the cube is equal to √(a^2 + b^2 + c^2) = √(3s^2) = s√3

So,

if the diagonals bisect each other, they will just bisect each other.

By using law of cosines in 4 triangles, and also vertical angles are equal, lets say,

<A = <B by vertical angles and also, <C = <D

By Law of cosines,

cos(A) = (b^2 + c - a^2)/(2bc) ----> cos (A) = cos (B) = -1/3

cos (C) = (a^2 + b^2 - c^2)/(2bc) ----> cos (C) = (cos D) = 1/3

therefore,

(cosA)^2 + (cosB)^2 + (cosC)^2 + (cosD)^2= (-1/3)^2 + (-1/3)^2 + (1/3)^2 + (1/3)^2 = 4(1/9) = 4/9