In a cubical hall A B C D P Q R S with
each side 10 m, $G$is the centre of the wall $B C R Q$ and $T$
is the mid point of the side $AB$. The angle of elevation of $G$
at the point $T$is

Let $H$be the mid point of$BC$since $\angle TBH$

$={90}^{\circ },T{H}^2=B{T}^2+B{H}^2=5^2+5^2=50$

Also since $\mathrm{\angle }THG={90}^{\circ },T{G}^2=T{H}^2+G{H}^2=50+25=75$

Let $\theta$ be the required angle of elevation of $G$at $T$

(Fig. 27.11)

then $\sin \theta =\frac{GH}{TG}=\frac{5}{5\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \mid \theta ={\sin }^{-1}(1/\sqrt{3})$