Question

A flag staff of height $10$ metres is placed on the top of a tower of height $30$ metres. At the top of a tower of height $40$ metres, the flag staff and the tower subtend equal angles then the distance between the two towers in meters is ....

Answer 1

Let $AB$ be the tower and $EA$ be the flag staff.

From the figure, it can be observed that $△EAC$ and $△EBC$ are right angled triangles.

$\begin{array}{c}\tan \theta =\frac{40}{BD}\\ BD=\frac{40}{\tan \theta }\\ Now,\\ \tan 2\theta =\frac{CD}{BD}=\frac{40}{BD}\\ BD=\frac{40}{\tan 2\theta }\\ \tan \theta =\frac{AC}{CE}=\frac{10}{CE}\\ \tan \theta =\frac{10}{CE}\\ CE=\frac{10}{\tan \theta }\\ CE=BD\\ \frac{10}{\tan \theta }=\frac{40}{\tan 2\theta }\\ \tan 2\theta =4\tan \theta \\ \frac{2\tan \theta }{1-{\tan }^2\theta }=4\tan \theta \\ 2\tan \theta =4\tan \theta -4{\tan }^3\theta \\ 4{\tan }^3\theta -2\tan \theta =0\\ 2\tan \theta \left(2{\tan }^2\theta -1\right)=0\\ \tan \theta =0\text{or}2{\tan }^2\theta -1=0\\ {\tan }^2\theta =\frac{1}{2}\\ \tan \theta =\frac{1}{\sqrt{2}}\\ Now,\\ CE=\frac{10}{\tan \theta }\\ BD=CE=\frac{10}{\tan \theta }\\ =\frac{10}{\frac{1}{\sqrt{2}}}\\ \therefore BD=CE=10\sqrt{2}m.\end{array}$