Question

From the top of a $10m$ high building, the angle of elevation of a tower is ${60}^{\circ }$ and the angle of depression of its foot at ${45}^{\circ }$. Determine the height of the tower.

Answer 1

The given scenario is shown in the attached figure.

Given:

$DE=10$

$\angle CDB={45}^o$

$\angle CDA={60}^o$

To find: $AB$

Solution:

$\tan \angle CDB=\frac{CB}{CD}$

$\Rightarrow \tan {45}^o=\frac{DE}{CD}\because CB=DE$

$1=\frac{10}{CD}$

$CD=10m$

$\tan \angle CDA=\frac{CA}{CD}$

$\Rightarrow \tan {60}^o=\frac{CA}{10}$

$\sqrt{3}=\frac{CA}{10}$

$CA=10\sqrt{3}m$

$AB=AC+BC$

$AB=10+10\sqrt{3}\approx 27.32m$

Hence, the height of the tower is $27.32m$