Question

A flagstaff $5$ m high is placed on a building $25$ m high. If the flag and building be subtend equal angels on the observer at a height $30$ m, then the distance between observer and the top of flag is

• A. $\frac{5\sqrt{3}}{2}$
• B. $5\sqrt{\frac{2}{3}}$
• C. $5\sqrt{\frac{3}{2}}$
• D. $\frac{5\sqrt{2}}{3}$

Answer 1

The correct option is C $5\sqrt{\frac{3}{2}}$

Given, $\angle DAE=\angle DAC=\theta$ (Let's assume)

Thus, $\angle EAC=2\theta$

Let's assume the distance between the building and the observer is ${}^{\prime }{x}^{\prime }m$.

Thus, for the triangle $\Delta ACE$;

$EC=ED+DC=5+25=30m$

Opposite/adjacent $=\frac{EC}{EA}=\frac{30}{x}=tan2\theta$

For the $\Delta EDA;$

Opposite/adjacent $=\frac{ED}{EA}=\frac{5}{x}=tan\theta$

Now, $tan2\theta =\frac{2tan\theta }{(1-ta{n}^2\theta )}$

or, $\frac{30}{x}=\frac{2\left(\frac{5}{x}\right)}{(1-\frac{25}{x^2})}$ [As $tan\theta =\frac{5}{x}$ and $tan2\theta =\frac{30}{x}]$

or, $3=\frac{1}{(1-\frac{25}{x^2}}$

or, $3x^2-75=x^2$

or, $2x^2=75$

or, $x=5\sqrt{\left(\frac{3}{2}\right)}=6.124m$

Thus the distance between the building and the observer is $5\sqrt{\left(\frac{3}{2}\right)}m$