Question

From the top of a building $60$ m high, the angles of depression of the top and bottom of a tower are observed to be ${45}^0$ and ${60}^0$ respectively. Then, find the height of the tower . [take, $\sqrt{3}=1.732$]

Answer 1

Let $AC$ be the tower and $BE$ be the building. Let height of the tower be hm.

It is given that the angles of depression of the top $C$ and bottom $A$ of the tower observe from top of the building be ${45}^o$ and ${60}^o$ respectively.

In right triangle $CDE$, we have

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

$\Rightarrow 1=\frac{60-h}{CD}[DE,BE-BD=BE-AC]$

$\Rightarrow CD=60-h-----\left(1\right)$

In right triangle $ABE$, we have

$\tan {60}^o=\frac{BE}{AB}$

$\sqrt{3}=\frac{60}{CD}(AB=CD)$

$CD=\frac{60}{\sqrt{3}}------\left(2\right)$

comparing $\left(1\right)$ and $\left(2\right)$

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

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

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

$\Rightarrow =\frac{60(\sqrt{3}-1)}{\sqrt{3}}=\frac{60.(1.732-1)}{1.732}$

$\therefore$ of the tower is $12.8m$