Question

From the foot and the top of a building of height $230m$, a person observes the top of a tower with angles of elevation of $\beta$ and $\alpha$ respectively. What is the distance between the top of the building and the top of the tower if $\tan \alpha =\frac{5}{12}$ and $\tan \beta =\frac{4}{5}$.

• A. $400m$
• B. $250m$
• C. $600m$
• D. $650m$

Answer 1

The correct option is D $650m$


Let ED be the building and AC be the tower.
Given that $ED=230m$, $\angle ADC=\beta$ and $\angle AEB=\alpha$.
Also, given:- $\tan \alpha =\frac{5}{12}$ and $\tan \beta =\frac{4}{5}$.

Let $AC=h$
Required distance, is the distance between the top of these buildings = AE.
From the right $\Delta ABE$,
$\tan \alpha =\frac{AB}{BE}$
$\Rightarrow \frac{5}{12}=\frac{h-230}{BE}$ [$\because$ AB = (AC-BC) = (AC-ED) = (h-230)]
$\Rightarrow BE=\frac{12\times (h-230)}{5}$ $\dots \left(1\right)$

From the right $\Delta ACD$,
$\tan \beta =\frac{AC}{CD}$
$\Rightarrow \frac{4}{5}=\frac{h}{CD}$[$\because$ AC = h]
$\Rightarrow CD=\frac{5h}{4}$ $\dots \left(2\right)$

From the diagram, BE = CD
$\Rightarrow \frac{12(h-230)}{5}=\frac{5h}{4}$ [from (1) and (2)]
$\Rightarrow 48h-(4\times 12\times 230)=25h$
$\Rightarrow 23h-(4\times 12\times 230)=0$
$\Rightarrow h=\frac{4\times 12\times 230}{23}=480m$ $\dots \left(3\right)$
$\therefore AB=(AC-BC)=480-230=250m$

In $\Delta ABE$, $\tan \alpha =\frac{5}{12}$.
$\therefore \sin \alpha =\frac{5}{13}$
$\Rightarrow \frac{AB}{AE}=\frac{5}{13}$
$\Rightarrow AE=AB\times \frac{13}{5}$
$\Rightarrow AE=250\times \frac{13}{5}=650m$

$\therefore$ Distance between the top of the buildings = $650m$