Question

From the top of a $38m$ high tower, the angle of depression of the top of a building is ${30}^o$. If the height of the building is $18m$, find the distance between the tower and the building.

Answer 1

Let, $AB$ = Height of the tower=$38m$

$CD$ = Height of the building=$18m$

From diagram, $BE=CD=18m$

$AE=AB-BE=(38-18)m=20m$

Let, the distance between tower and building be,$BD=CE=xm$

From $\Delta AEC$

$\tan {30}^o=\frac{AE}{EC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{x}$
$\Rightarrow x=20\sqrt{3}$
$\Rightarrow x=20\times 1.732$

$\Rightarrow x=34.64$

Hence, the distance between tower and building is $34.64m$