Let the distance, be $\mathrm{x}$, and angle $\mathrm{APB}=\theta$, then $\angle \mathrm{BPC}=\angle \mathrm{APC}=\theta / 2$
In triangle $\Delta \mathrm{APB}$,
$\tan \theta=\frac{\mathrm{AB}}{\mathrm{x}}=\frac{50}{\mathrm{x}}$
.(1)
and in triangle $\mathrm{APC}$
$\tan \frac{\theta}{2}=\frac{\mathrm{AC}}{\mathrm{x}}=\frac{24}{\mathrm{x}}$
$\tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}$
Putting the value of $\tan \theta$ and $\tan \frac{\theta}{2}$
From equation (1) and (2) in equation (3),
$\frac{50}{x}=\frac{2 \times \frac{24}{x}}{1-\left(\frac{24}{x}\right)^{2}}$
or, $\frac{50}{x}=\frac{48}{x} \times \frac{x^{2}}{x^{2}-(24)^{2}}$
or, $50\left\{\mathrm{x}^{2}-(24)^{2}\right\}=48 \mathrm{x}^{2}$
or, $50 \mathrm{x}^{2}-50 \times(24)^{2}=48 \mathrm{x}^{2}$
or, $2 \mathrm{x}^{2}=(24)^{2} \times 50$
$\mathrm{x}=25 \times(24)^{2}$
$\mathrm{x}=|5 \times 24|=120 \mathrm{~m}$
Such a point can exist on the either side of the tower.