Question

There are two windows in a house. A window is at a height of 2 m above the ground and the other window is 3 m vertically above the lower window. A balloon is flying near the house. At an instant, angles of elevation of the balloon from these windows are observed as ${45}^{\circ }and{30}^{\circ }$ from lower and upper window respectively. Find the height of the balloon from the ground.

Answer 1

Let H be the height of the balloon from the ground and C and D be the position of the windows.

At C and D, angle of elevation are $\angle ECG={45}^{\circ }and\angle FDG={30}^{\circ }$
Let CE = DF = x m and FG = h m

In $\Delta CEG$, we have
$tan{45}^{\circ }=\frac{EG}{EC}\Rightarrow 1=\frac{EF+FG}{EC}\Rightarrow 1=\frac{3+h}{x}$
$\Rightarrow x=3+h\dots \left(i\right)$

In $\Delta DFG$, we have
$tan{30}^{\circ }=\frac{GF}{DF}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\Rightarrow x=\sqrt{3}h\dots \left(ii\right)$

Substituting $x=\sqrt{3}h$ in (i), we get
$\sqrt{3}h=3+h$
$\Rightarrow \sqrt{3}h-h=3\Rightarrow h(\sqrt{3}-1)=3\Rightarrow h=\frac{3}{\sqrt{3}-1}\Rightarrow h=\frac{3}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\Rightarrow h=\frac{3(\sqrt{3}+1)}{3-1}\Rightarrow h=\frac{3\times (1.732+1)}{2}\Rightarrow h=\frac{3\times 2.732}{2}\Rightarrow h=4.098m$

Hence, the height of the balloon from the ground is
H = EA + FE + h
$\Rightarrow$ H = 2 + 3 + 4.098 $\Rightarrow$ H = 9.098 m