Question : The angles of elevation of an aeroplane flying vertically above the ground, as observed from the two consecutive stones,1 km apart, are 45° and 60°. The height of the aeroplane above the ground in km is:
Option 1: $(\sqrt{3}+1)$
Option 2: $(\sqrt{3}+3)$
Option 3: $\frac{1}{2}(\sqrt{3}+1)$
Option 4: $\frac{1}{2}(\sqrt{3}+3)$
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Correct Answer: $\frac{1}{2}(\sqrt{3}+3)$
Solution :
Let A be the position of the aeroplane and B and C be the two milestones.
So, BC = 1 km
$\angle \text{ABD} =$ 45° and $\angle \text{ACD} =$ 60°
Let AD = $h$ km and CD = $x$ km
Now, $\tan 45° = \frac{h}{x+1}$
⇒ $1 = \frac{h}{x+1}$
⇒ $x = h-1$
Also, $\tan 60° = \frac{h}{x}$
⇒ $\sqrt{3} = \frac{h}{x}$
⇒ $x = \frac{h}{\sqrt{3}}$
So, $\frac{h}{\sqrt{3}} = h-1$
⇒ $h = \frac{\sqrt{3}}{\sqrt{3}-1}$
Rationalising by ($\sqrt{3}+1$) we get,
⇒ $h = \frac{1}{2}(3+\sqrt{3})$ km
Hence, the correct answer is $\frac{1}{2}(\sqrt{3}+3)$.
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