From an aeroplane flying, vertically above a horizontal road, the angles of depression of two consecutive stones on the same side of the aeroplane are observed to be $30^{\circ}$ and $60^{\circ}$ respectively. The height at which the aeroplane is flying in $\mathrm{km}$, is

• A

  2
• B

  $\frac{2}{\sqrt{3}}$
• Correct answer
  C

  $\frac{\sqrt{3}}{2}$
• D

  $\frac{4}{\sqrt{3}}$

Solution

Let the distance of two consecutive stones are $(x, x+1)$. In $\triangle B C D$, we have

$\tan 60^{\circ}=\frac{h}{x}$ $\Rightarrow \quad x=\frac{h}{\sqrt{3}}...(i)$ In $\triangle A B C$, we have $\tan 30^{\circ}=\frac{h}{x+1}$ $\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x+1}$ $\Rightarrow \quad \frac{h}{\sqrt{3}}+1=\sqrt{3 h}$ [by Eq. (i)] $\begin{array}{ll}\Rightarrow & h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=1 \\ \Rightarrow & h \cdot\left(\frac{2}{\sqrt{3}}\right)=1 \Rightarrow h=\frac{\sqrt{3}}{2}\end{array}$

Hence, the correct answer is option: 'C'.