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If $\theta$ is the angle, in degrees, between the longest diagonal of the cube and any one of the edges of the cube, then, $\cos \theta =$

1. $\frac{1}{2} \\$
2. $\frac{1}{\sqrt{3}} \\$
3. $\frac{1}{\sqrt{2}} \\$
4. $\frac{\sqrt{3}}{2}$

### 1 comment

see this video to understand longest diagonal of cube.

Option B

The longest diagonal would be from one corner vertex to the diagonally opposite corner vertex.

$\text{Length of diagonal of a side} = \sqrt{a^2 + a^2} = \sqrt{2}a$

The diagonal of a square face of cube, a side of the cube and the longest diagonal will form a right angled triangle with longest diagonal as the hypotenuse.

$\therefore \text{Length of the longest diagonal} = \sqrt{a^2 + \left(\sqrt{2}a\right)^2} = \sqrt{3}a$

$\cos\theta = \frac{\text{Base}}{\text{Hypotenues}} = \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt{3}}$
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√2/√3 is possible not √3/√2  but the condition should be that θ is the angle between two diagonals
The length of the longest diagonal is coming as √3a / 2 when considering the triangle consisting theta and the base is simply a.

So cos theta should give = a / ( √3a/2 )

Where I'm doing wrong ?

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