# Solved Problem 14 | Rectangular Parallelepiped

**Problem 14**

Find the angles that the diagonals of the rectangular parallelepiped 2 in. by 3 in. by 4 in. makes with the faces.

**Solution 14**

Angle made by the diagonal to the 4 in. by 3 in. faces

$\tan \alpha = \dfrac{2}{x}$

$\tan \alpha = \dfrac{2}{x}$

$\tan \alpha = \dfrac{2}{\sqrt{4^2 + 3^2}}$

$\alpha = 21^\circ 48'$ *answer*

Angle made by the diagonal to the 4 in. by 2 in. faces

$\tan \beta = \dfrac{3}{y}$

$\tan \beta = \dfrac{3}{\sqrt{4^2 + 2^2}}$

$\beta = 33^\circ 51'$ *answer*

Angle made by the diagonal to the 3 in. by 2 in. faces

$\tan \theta = \dfrac{4}{z}$

$\tan \theta = \dfrac{4}{\sqrt{3^2 + 2^2}}$

$\theta = 47^\circ 58'$ *answer*

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