Question

A pole $50mt$ high stand on a building of $250mt$ high. To an observer at a height of $300mt$ the building and the pole subtends same angle. The distance of the top of pole is

• A. $25mt$
• B. $25\surd 3mt$
• C. $50mt$
• D. $25\surd 6mt$

Answer 1

The correct option is D $25\surd 6mt$

Consider the attached figure while going through the following steps.

Let the distance be $"x"$

so, we have,

$tan\varphi =\frac{50}{x}$

and

$tan2\varphi =\frac{(250+50)}{x}$

$\frac{2tan\varphi }{(1-ta{n}^2\varphi )}=\frac{300}{x}$

$\frac{\left[2\right(\frac{50}{x}\left)\right]}{[1-(\frac{50}{x}{)}^2]}=\frac{300}{x}$

$\frac{100}{x}=\frac{300}{x}(1-\frac{2500}{x^2})$

$1-\frac{2500}{x^2}=\frac{1}{3}$

$\frac{2500}{x^2}=\frac{2}{3}$

$x^2=3750$

$x=25\sqrt{6}m$

Therefore, the distance of the observer from top of the pole is $25\sqrt{6}m$