Question

A window in a building is at a height of 10 m from the ground. The angle of depression of a point P on the ground from the window is 30°. The angle of elevation of the top of the building from the point P is 60°. Find the height of the building.

Or, in a violent storm, a tree is bent by the wind. The top of the tree meets the ground at an angle of 30°, at a distance of 30 metres from the root. At what height from the bottom did the tree get bent? What was the original height of the tree?

Answer 1

Given, height of the window from the ground DB = 10 m
Let H be the height of the building.
Let O be the point of observation. Then,

DB = 10 m, $\angle$DPB = 30° and $\angle$APB = 60°
BP = x m
$\tan 30°=\frac{10}{x}$
$\frac{1}{\sqrt{3}}=\frac{10}{x}\Rightarrow x=10\sqrt{3}\mathrm{m}$ ...............(i)
Again,
$\tan 60°=\frac{H}{x}$
$\Rightarrow \sqrt{3}=\frac{H}{x}$
$\Rightarrow H=\sqrt{3}x$
$\Rightarrow H=\left(\sqrt{3}\times 10\sqrt{3}\right)\mathrm{m}=30\mathrm{m}$
Hence, required height of the building is 30 m.

OR
Let AB be the original height of the tree, which gets bent at a point C. After being bent, let the part CB take the position CD, meeting the ground at D. Then AD = 30 m, ∠ADC = 30°, ∠DAC = 90° and CD = CB

Let AC = x metres and CD = CB = y metres
From right-angled ΔDAC, we have:
$\frac{AC}{AD}=\tan 30°=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{x}{30}=\frac{1}{\sqrt{3}}$
$\Rightarrow x=\left(30\times \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\right)=10\sqrt{3}\mathrm{m}$
Also, from right-angled ΔDAC, we have:
$\frac{DC}{AD}=\sec 30°=\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{y}{30}=\frac{2}{\sqrt{3}}$
$\Rightarrow y=\left(30\times \frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\right)=20\sqrt{3}\mathrm{m}$
Thus AB = (x + y) metres = $\left(10\sqrt{3}+20\sqrt{3}\right)\mathrm{m}=30\sqrt{3}\mathrm{m}$
Hence, the tree got bent at a height of $10\sqrt{3}\mathrm{m}$ and the original height of the tree was $30\sqrt{3}\mathrm{m}$.