It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60°. The height of the chimney is
(a) $3 \sqrt{2} x$
(b) $2 \sqrt{3} x$
(c) $\frac{\sqrt{3}}{2} x$
(d) $\frac{2}{\sqrt{3}} x$
Let be the height of chimney
Given that: angle of elevation changes from angle $\angle D=30^{\circ}$ to $\angle C=60^{\circ}$.
Then Distance becomes $C D=x$ and we assume $B C=y$
Here, we have to find the height of chimeny.
So we use trigonometric ratios.
In a triangle $A B C$,
$\Rightarrow \tan C=\frac{A B}{B C}$
$\Rightarrow \tan 60^{\circ}=\frac{A B}{B C}$
$\Rightarrow \sqrt{3}=\frac{h}{y}$
$\Rightarrow y=\frac{h}{\sqrt{3}}$
Again in a triangle ABD,
$\Rightarrow \tan D=\frac{A B}{B C+C D}$
$\Rightarrow \tan 30^{\circ}=\frac{h}{y+x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{y+x}$
$\Rightarrow \sqrt{3} h=y+x$
$\Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+x \quad\left[\right.$ Put $\left.y=\frac{h}{\sqrt{3}}\right]$
$\Rightarrow h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=x$
$\Rightarrow h=\frac{x}{\sqrt{3}-\frac{1}{\sqrt{3}}}$
$\Rightarrow h=\frac{\sqrt{3} x}{2}$
Hence the correct option is $c$.