Question

A flagstaff 5 meter high stands on a building 25 meter high. An observer is at a height of 30 meters. The flag and the building subtend equal angles. The distance of the observes from the top of the flagstaff is
A.$ \dfrac{{5\sqrt 3 }}{{\sqrt 8 }} $
B.$ \dfrac{{5\sqrt 3 }}{{\sqrt 2 }} $
C.$ \dfrac{{4\sqrt 3 }}{{\sqrt 2 }} $
D.$ \dfrac{{7\sqrt 3 }}{{\sqrt 2 }} $

Answer 1

Hint: Draw the figure that defines the given data and then apply the trigonometric ratio rule on triangle OBC and then use the result and a formula given below, to find the desired result:
$ \tan 2\alpha = \dfrac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }} $

Complete step-by-step solution:
We have given that A flagstaff of height 5 meters stands on a building whose height is 25 meters high. It is also given that there is an observer at the height of 30 meters and both the flag and the building are subtended at equal angles with the observer.
Assume AB is the building and BC is the flag, O is the observer who stands at the height of 30 meters and his distance from the top of the flag be $x$. Also assume that the equal subtended angle from the flag and the building be$ \alpha $. Then the figure- defining the environment is given as:

First take a look on the triangle $OBC$, using the trigonometric ratio, the tan of the $ \alpha $ is the ratio of the opposite side to the adjacent side of the triangle, so we have
$ \tan \alpha = \dfrac{{{\text{Opposite side}}}}{{{\text{Adjacent side}}}} $
We have the opposite side as $BC$ and the adjacent side as $ OC $, whose values are:
$ BC = 5 $ and $ OC = x $.
Substitute these values in the trigonometric ratio, we have
$ \tan \alpha = \dfrac{{BC}}{{OC}} $
$ \Rightarrow \tan \alpha = \dfrac{5}{x} $
Now, take a look at the triangle $ AOC $
Using the trigonometric ratio, the tan of the angle $ 2\alpha $ is given as the ratio of the opposite side and the adjacent side,
$ \tan 2\alpha = \dfrac{{{\text{Opposite side}}}}{{{\text{Adjacent side}}}} $
In this triangle, the opposite side is $ AC $ and the adjacent side is $ OC $, whose values are:
$ AC = 30 $ and $ OC = x $
Substitute these values in the trigonometric ratio gives,
$ \tan 2\alpha = \dfrac{{{\text{AC}}}}{{OC}} $
$ \tan 2\alpha = \dfrac{{30}}{x} $
We know that,
$ \tan 2\alpha = \dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} $
So, the above equation becomes:
$ \dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \dfrac{{30}}{x} $
Substitute the value $\tan \alpha = \dfrac{5}{x}$ in the equation:
$ \dfrac{{2\left( {\dfrac{5}{x}} \right)}}{{1 - {{\left( {\dfrac{5}{x}} \right)}^2}}} = \dfrac{{30}}{x} $
Now, solve the equation for the value of$x$,
$ \dfrac{{\dfrac{{10}}{x}}}{{1 - \left( {\dfrac{{25}}{{{x^2}}}} \right)}} = \dfrac{{30}}{x} $
$ \Rightarrow \dfrac{{\dfrac{{10}}{x}}}{{\left( {\dfrac{{{x^2} - 25}}{{{x^2}}}} \right)}} = \dfrac{{30}}{x} $
$ \Rightarrow \dfrac{x}{{{x^2} - 25}} = \dfrac{3}{x} $
Perform a cross multiplication and solve the equation for the value of $x$.
$ \Rightarrow \dfrac{x}{{{x^2} - 25}} = \dfrac{3}{x} $
$ \Rightarrow 3{x^2} - 75 = {x^2} $
$ \Rightarrow 2{x^2} = 75 $
$ \Rightarrow {x^2} = \dfrac{{75}}{2} $
$ \Rightarrow x = 5\sqrt {\dfrac{3}{2}} $
Therefore, the distance of the observer from the top of the flag is $ 5\sqrt {\dfrac{3}{2}} $.

Hence,option(B) is the correct option.

Note: After applying the trigonometric ratio, don’t try to find the value of $x$, find the value of $ \tan \alpha $, and then use that value of $ \tan \alpha $ for further calculation.