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 angles. The horizontal distance
of the observer from the pole is

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$25 m$

$50 m$

$25 \sqrt{6} m$

$25 \sqrt{3} m$

answer is C.

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Detailed Solution

Let $P Q$ be the pole on the building $Q R$ and
$O$ be the observer.

Then $P Q = 50, Q R = 250 (Fig. 27.20)$

$\Rightarrow P R = 300$ so the observer is at the same height as the
top $P$ of the pole. Let $O P = x .$ Then from right angled
triangles $O P Q$ and $O P R,$

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

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

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