Question : From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60° and 30°, respectively. The height (in km) of the aeroplane from the road at that instant, is:
Option 1: $\frac{\sqrt{3}}{2}$
Option 2: $\frac{\sqrt{3}}{3}$
Option 3: $\frac{\sqrt{3}}{4}$
Option 4: $\sqrt{3}$
Correct Answer: $\frac{\sqrt{3}}{4}$
Solution :
Let the aeroplane be at point A at a height $h$ above ground level.
Let B and C be two consecutive kilometre stones with angles of depression $60°$ and $30°$, respectively.
Now, $\tan 60°=\frac{h}{BD}$ and $\tan 30°=\frac{h}{CD}$
⇒ $BD=h\cot 60°$ and $CD=h\cot 30°$
$\because BD+CD=1$
$\therefore h\cot 60°+h\cot 30°=1$
⇒ $h(\frac{1}{\sqrt{3}}+\sqrt{3})=1$
⇒ $h=\frac{\sqrt{3}}4$
Hence, the correct answer is $\frac{\sqrt{3}}4$ km.
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