Aspire Faculty ID #11983 · Topic: NIMCET 2025 · Just now
NIMCET 2025

An airplane, when 4000m high from the ground, passes vertically above another airplane at an instant when the angles of elevation of the two airplanes from the same point on the ground are 60° and 30°, respectively. Find the vertical distance between the two airplanes.

Solution

Let the higher airplane be at a height of $4000\,\text{m}$. 
From a point on the ground, the angles of elevation to the two airplanes are $60^\circ$ (upper plane) and $30^\circ$ (lower plane). 

Let the horizontal distance from the observer to the airplanes be $x$. 

For the upper airplane: 
$\tan 60^\circ = \dfrac{4000}{x}$ 
$\sqrt{3} = \dfrac{4000}{x}$ 
$\Rightarrow x = \dfrac{4000}{\sqrt{3}}$. 
 For the lower airplane with height $h$: 
$\tan 30^\circ = \dfrac{h}{x}$ 
$\dfrac{1}{\sqrt{3}} = \dfrac{h}{4000/\sqrt{3}}$ 
 Thus, $h = \dfrac{4000}{3}$. 
 Now the vertical distance between the two airplanes: $4000 - \dfrac{4000}{3} = \dfrac{8000}{3}.$ 
 Final Answer: $\displaystyle \frac{8000}{3}\text{ m}$.
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