Question

ABCDEFGH is a cuboid. Find the angle between AG and plane EFGH(angle AGE).

Answer 1

In $\Delta$ AEG

$\tan \left(\angle AGE\right)=\frac{AE}{EG}=\frac{8}{\sqrt{137}}=0.6837$

$\angle AGE={\tan }^{-1}\left(0.6837\right)$

$={34.36}^o$.