Consider a cubical hall ABCDPQRS with each side 20m and base as...

Let $H$ be the midpoint of $BC$

Since $∠TBH = {90}^{o},$

$⇒ TH^{2} = BT^{2} + BH^{2} = {10}^{2} + {10}^{2} = 200$
Also, since $∠THG = {90}^{o},$

$⇒ TG^{2} = TH^{2} + GH^{2} = 200 + 100 = 300$
Let $\theta$ be the required angle of elevation of $G$ at $T$.
Then, $sin \theta = \dfrac{GH}{TG} = \dfrac{10}{10\sqrt{3}} = \dfrac{1}{\sqrt{3}}$
$⇒ \theta = \sin^{-1} \dfrac{1}{3}$