The angle of elevation of the top of a tower from two points at distances $s$ and $t$ from its foot are complementary. Prove that the height of the tower is $\sqrt{s t}$.

(N/A) Let the height of the tower be $h$ and the tower be represented by $AC$,where $C$ is the foot of the tower.
Let the two points be $B$ and $P$ on the same line such that $BC = s$ and $PC = t$.
Let $\angle ABC = \theta$. Since the angles of elevation are complementary,$\angle APC = 90^{\circ} - \theta$.
In $\triangle ABC$,$\tan \theta = \frac{AC}{BC} = \frac{h}{s}$ --- $(i)$
In $\triangle APC$,$\tan(90^{\circ} - \theta) = \frac{AC}{PC} = \frac{h}{t}$.
Since $\tan(90^{\circ} - \theta) = \cot \theta$,we have $\cot \theta = \frac{h}{t}$.
Therefore,$\frac{1}{\tan \theta} = \frac{h}{t} \implies \tan \theta = \frac{t}{h}$ --- $(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \frac{h}{s} \cdot \frac{h}{t}$
$1 = \frac{h^2}{st}$
$h^2 = st$
$h = \sqrt{st}$
Thus,the height of the tower is $\sqrt{st}$.