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A man observes the elevation of a balloon to be $30^{\circ} . \mathrm{He}$, then walks $1 \mathrm{~km}$ towards the balloon and finds that the elevation is $60^{\circ}$. What is the height of the balloon?
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Verified Answer
The correct answer is:
$\sqrt{3} / 2 \mathrm{~km}$
Let the height of the balloon be $\mathrm{h}$ and new distance between $\mathrm{BC}$ be y as shown in figure given below.
$\tan 60^{\circ}=\frac{C D}{B C}$
$\Rightarrow \sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y$
and now in $\triangle \mathrm{ADC}, \quad \tan 30^{\circ}=\frac{C D}{A C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{1+y}$
$\Rightarrow 1+y=h \sqrt{3} \Rightarrow 1+y=3 y$
$\Rightarrow y=\frac{1}{2}$
$\therefore \quad h=\frac{\sqrt{3}}{2}$
$\tan 60^{\circ}=\frac{C D}{B C}$
$\Rightarrow \sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y$
and now in $\triangle \mathrm{ADC}, \quad \tan 30^{\circ}=\frac{C D}{A C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{1+y}$
$\Rightarrow 1+y=h \sqrt{3} \Rightarrow 1+y=3 y$
$\Rightarrow y=\frac{1}{2}$
$\therefore \quad h=\frac{\sqrt{3}}{2}$
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