Let $H$ be the midpoint of $BC$

Since $\angle TBH={90}^{\mathrm{o}},$

$\Rightarrow T{H}^2=B{T}^2+B{H}^2={10}^2+{10}^2=200$
Also, since $\angle THG={90}^{\mathrm{o}},$

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