Question

Find the acute angle formed by two diagonals of a cube .​

Answer 1

Given :

A cube

To find :

Acute angle formed by two diagonals of a cube.

Solution :

​Let us assume that one of the vertices is at the origin.

Let the length of side of the cube be 'a'.

So, consider the diagonal d₁ through vertices (0 , 0 , 0) and (a , a , a)

Consider the diagonal d₂ through vertices (0 , a , 0) and (a , 0 , a)

d₁ = (a - 0 , a - 0 , a - 0)

d₁ = (a , a , a)

Similarly, d₂ = (a - 0 , 0 - a , a - 0)

d₂ = (a , -a , a)

$cos \theta = \frac{l_1 l_2 + m_1m_2 +n_1n_2}{\sqrt{l_1^{2} + m_1^{2} + n_1^{2} } \sqrt{l_2^{2} + m_2^{2} + n_2^{2}} }$

$cos \theta = \frac{a^{2}-a^{2}+a^{2} }{\sqrt{a^{2}+a^{2}+a^{2}} \sqrt{a^{2}+a^{2}+a^{2}} }$

$cos\theta = \frac{a^{2} }{\sqrt{3}a.\sqrt{3} a }$

cos θ = 1 / 3

θ = cos⁻¹(1 / 3)

θ = 70.52°

The acute angle formed by two diagonals of a cube is 70.52°

Answer 2

Answer:

Ans. arc cos 1/3 or 70°32`

Step-by-step explanation: