Question

A flagstaff is placed on top of a building. The flagstaff and the building subtend equal angles at a point on level ground which is $200m$ away from the foot of the building. If the height of the flagstaff is $50m$ and the height of the building is $h$, which of the following is true?

• A. $h^3-50h^2+(200{)}^{2}h+50(200{)}^2=0$
• B. $h^3+50h^2+(200{)}^{2}h-50(200{)}^2=0$
• C. $h^3-50h^2-(200{)}^{2}h+50(200{)}^2=0$
• D. none of these

Answer 1

The correct option is B $h^3+50h^2+(200{)}^{2}h-50(200{)}^2=0$


Let AD be the flagstaff and CD be the building.
Given that the flagstaff and the building subtend equal angles at point B.
Given that $AD=50m$, $CD=h$, and $BC=200m$.
Let, $\angle ABD=\theta$
$\therefore$ $\angle DBC=\theta$
Then $\angle ABC=2\theta$

From the right $\Delta BCD$
$\tan \theta =\frac{DC}{BC}=\frac{h}{200}\dots \left(1\right)$
From the right $\Delta BCA$
$\tan 2\theta =\frac{AC}{BC}=\frac{AD+DC}{200}=\frac{50+h}{200}\dots \left(2\right)$
$\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{50+h}{200}$ [$\because$ $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$]
$\frac{\frac{2h}{200}}{1-\frac{h^2}{{200}^2}}=\frac{50+h}{200}$ [substituted value of $\tan \theta$ from equation 1]
$\Rightarrow 2h=(1-\frac{h^2}{{200}^2})\times (50+h)$
$\Rightarrow 2h=50+h-\frac{50h^2}{{200}^2}-\frac{h^3}{{200}^2}$
$\Rightarrow 2\left({200}^2\right)h=50(200{)}^2+(200{)}^{2}h-50h^2-h^3$
$\Rightarrow h^3+50h^2+(200{)}^{2}h-50(200{)}^2=0$