It is found that on walking x meters towards a chimney in a horizontal

Question:

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) $\frac{2}{\sqrt{3}} x$

Solution:

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 $c$.

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