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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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