Question

A pole of $50$ 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. The distance of the observer from the top of the pole is:-

Answer 1

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

Then $PQ=50m,QR=250m$

$PR=300$ so the observer is at the same height as the top $P$ of the pole.

Let $OP=x$.

Then from right angled triangles $OPQ$ and $OPR$, we write :

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

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

$\frac{2\times \frac{50}{x}}{1-(\frac{50}{x}{)}^2}=\frac{300}{x}$

$\{1-{\left(\frac{50}{x}\right)}^2\}=\frac{1}{3}$

${\left(\frac{50}{x}\right)}^2=\frac{2}{3}$

$\Rightarrow x=\frac{50\sqrt{3}}{\sqrt{2}}=25\sqrt{6}$.