Question

The horizontal distance between two towers is $140m$. The angle of elevation of the top of first tower when seen from the top of second tower is ${30}^0$. If the height of the second tower is $60m$ then find the height of first tower.

Answer 1

Let $AB$ is a tower of height $h$ and other tower is $ED$ of height $60m$.

Distance between two towers $EB=DC=140m$

The angle of elevation of the top of first tower when seen from the top of second tower is ${30}^0$.

So, $\angle ADC={30}^0$.

From right angled $\Delta ACD$

$\tan {30}^0=\frac{AC}{DC}$

$\frac{1}{\sqrt{3}}=\frac{h-60}{140}$

$h-60=\frac{140}{\sqrt{3}}$

$h=60+\frac{140}{\sqrt{3}}$

$=60+\frac{140}{\sqrt{3}}$

$=60+\frac{140}{1.732}$

$=140.8m$

Hence, height of first tower = $140.8m$.