Question

The shadow of a tower standing on a level ground is found to be $40$ m longer when the Sun's altitude is ${30}^{\circ }$ than when it is ${60}^{\circ }$. Find the height of the tower.

Answer 1

Given tower be $AB$

$\angle AACB=60$

and length of shadow=$BC$

$\angle ADB=30$

and length of shadow=$db$

Shadow is $40m$ when angle change from $60$ to $30$

$CD=40m$

We need to find height of tower i.e., $AB$

Since tower is vertical to ground

$\angle ABC=90$

In $△ABC$

$=>tan60=\frac{AB}{CB}$

$=>\sqrt{3}=\frac{AB}{CB}$

$=>CB=\frac{AB}{\sqrt{3}}------\left(i\right)$

In $△ABD$

$=>tan30=\frac{AB}{DB}$

$=>\frac{1}{\sqrt{3}}=\frac{AB}{DB}$

$=>DB=\sqrt{3}AB$

$=>DC+CB=\sqrt{3}AB$

$=>CB=\sqrt{3}AB-40----\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$ we get

$=>\frac{AB}{\sqrt{3}}=\sqrt{3}AB-40$

$=>AB=\sqrt{3}\left(\sqrt{3}AB\right)-40\sqrt{3}$

$=>2AB=40\sqrt{3}$

$=>AB=20\sqrt{3}$

Hence height of the tower=$AB=20\sqrt{3}$