Question

From the top of a building, 60 metres high, the angle of a depression of the top and bottom of a tower are observed to be ${30}^0$ and ${60}^0$, find the height of the tower.

Answer 1

Consider the problem

Consider $AC$ be the tower and $BE$ be the building

Let height of the tower be $hm$

And, it is given that the angle of depression of the top $C$ and bottom $A$ of the tower observed from top of the building be ${30}^{\circ }and{60}^{\circ }$

In right triangle $CDE$

$\tan {30}^{\circ }=\frac{DE}{CD}$

$\begin{array}{c}\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{CD}\left[DE=BE-BD=BE-AC\right]\\ \Rightarrow CD=\sqrt{3}\left(60-h\right)........\left(1\right)\end{array}$

And,

In right triangle $ABE$

$\begin{array}{c}\tan {60}^{\circ }=\frac{BE}{AB}\\ \Rightarrow \sqrt{3}=\frac{60}{CD}\left(AB=CD\right)\\ \Rightarrow CD=\frac{60}{\sqrt{3}}........\left(2\right)\end{array}$

Now,Comparing $\left(1\right)$ and $\left(2\right)$

$\begin{array}{c}\sqrt{3}\left(60-h\right)=\frac{60}{\sqrt{3}}\\ \Rightarrow 3\left(60-h\right)=60\\ \Rightarrow 180-3h=60\\ \Rightarrow 3h=120\\ \Rightarrow h=40\end{array}$

Hence, the height of the tower is $40m$