A pole 50 m high stands on a building 250 m high. To an observer at a height of$300 m,$the  building and the pole subtend equal angle . If the horizontal distance of the observer from the pole is x, then

Let $PQ$ be the pole on the building $QR$ and $O$ be the observer.

Then $PQ = 50, QR = 250$ (Fig. 12.28)

fi $PR = 300$ so the observer is at the same height as the top $P$ of the pole. So $OP = x.$ Then from right angled triangles $OPQ$ and $OPR,$

$\tan \theta =\frac{50}{x}$ and $\tan 2\theta =\frac{300}{x}.$

so that $\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{300}{x}$

$\begin{array}{l}\Rightarrow \frac{2\times \frac{50}{x}}{1-{\left(\frac{50}{x}\right)}^2}=\frac{300}{x}\Rightarrow 3\left\{1-{\left(\frac{50}{x}\right)}^2\right\}=1\\ \Rightarrow {\left(\frac{50}{x}\right)}^2=\frac{2}{3}\Rightarrow x=\frac{50\sqrt{3}}{\sqrt{2}}=25\sqrt{6}\end{array}$

and $\tan \theta =\frac{50}{25\sqrt{6}}=\sqrt{\frac{2}{3}}.$