Question

From an aero plane flying, vertically above a horizontal road, the angles of depression of two consecutive stones at a distance of $1$ km on the same side of the aeroplane are observed to be $30°$ and $60°$, respectively.

The height (in km) at which the aero plane is flying, is:

• A. $\frac{4}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{3}}$
• D. $2$

Answer 1

The correct option is B

$\frac{\sqrt{3}}{2}$

The explanation for the correct option:

Step: 1 Find the height of the triangle $DBC$ using trigonometric identities:

$$\begin{gathered} \tan \left(60°\right)=\frac{h}{x} \\ \Rightarrow h=x\sqrt{3}........\left(i\right) \end{gathered}$$

Step: 2 Find the height of the triangle $ABC$ using trigonometric identities:

$$\begin{gathered} \tan \left(30°\right)=\frac{h}{x+1} \\ \Rightarrow \sqrt{3}h=(x+1) \\ \Rightarrow h=\frac{(x+1)}{\sqrt{3}}..........\left(ii\right) \end{gathered}$$

Step: 3 Calculate the value of $h$:

From(i) and (ii) we get

$$\begin{gathered} x\sqrt{3}=\frac{(x+1)}{\sqrt{3}} \\ \Rightarrow 3x=x+1 \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$

Now, putting the value of $x$in any equation of $h$, we get:

$$\begin{gathered} h=\sqrt{3}x \\ \Rightarrow h=\frac{\sqrt{3}}{2} \end{gathered}$$

Hence, the height at which aero plane is flying is $\frac{\sqrt{3}}{2}$km.

Hence, Option(B) is the correct answer.