Solved Problem 14 | Rectangular Parallelepiped

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

$\alpha = 21^\circ 48'$           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