Question

From the top of a $7\mathrm{m}$ high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower.

Answer 1

Step 1: Find the horizontal distance between the tower and the building.

Let us assume that $\mathrm{AB}=7\mathrm{m}$ be the height of building, $\mathrm{EC}$ be the height of the tower.

$\mathrm{A}$ is the point from where the angle of elevation of tower is $60°$ and the angle of depression of its foot is $45°$

From the figure, $\mathrm{EC}=\mathrm{DE}+\mathrm{CD}$

Also, $\mathrm{CD}=\mathrm{AB}=7\mathrm{m}$ and $\mathrm{BC}=\mathrm{AD}$

In right triangle $\mathrm{ABC}$, $\tan \left(45°\right)=\frac{\mathrm{AB}}{\mathrm{BC}}$

$\Rightarrow$ $1=\frac{7}{\mathrm{BC}}$

$\Rightarrow$ $\mathrm{BC}=7\mathrm{m}$

$\Rightarrow$ $\mathrm{AD}=7\mathrm{m}$

Step 2: Find the height of the tower.

In right triangle $\mathrm{ADE},\tan \left(60°\right)=\frac{\mathrm{DE}}{\mathrm{AD}}$

$\Rightarrow$ $\sqrt{3}=\frac{\mathrm{DE}}{7}$

$\Rightarrow$ $\mathrm{DE}=7\sqrt{3}\mathrm{m}$

So, $\mathrm{EC}=\mathrm{DE}+\mathrm{CD}$

$\Rightarrow$$\mathrm{EC}=7\sqrt{3}+7$

$\Rightarrow$$\mathrm{EC}=7\left(1+\sqrt{3}\right)\mathrm{m}$

Hence, the height of the tower is $7\left(1+\sqrt{3}\right)\mathrm{m}.$