# The length of the shadow of a tower standing on level ground is found

Question:

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is

(a) $(\sqrt{3}+1) x$

(b) $(\sqrt{3}-1) x$

(c) $2 \sqrt{3} x$

(d) $3 \sqrt{2} x$

Solution:

Let  be the height of tower

Given that: angle of elevation of sun are $\angle D=30^{\circ}$ and $\angle C=45^{\circ}$.

Then Distance $C D=2 x$ and we assume $B C=X$

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle $A B C$,

$\Rightarrow \tan C=\frac{A B}{B C}$

$\Rightarrow \tan 45^{\circ}=\frac{A B}{B C}$

$\Rightarrow 1=\frac{h}{X}$

$\Rightarrow X=h$

Again in a triangle ABD,

$\Rightarrow \tan D=\frac{A B}{B C+C D}$

$\Rightarrow \tan 30^{\circ}=\frac{h}{X+2 x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{h+2 x} \quad[X=h]$

$\Rightarrow \sqrt{3} h=h+2 x$

$h(\sqrt{3}-1)=2 x$

$\Rightarrow h=\frac{2 x}{\sqrt{3}-1}$

$\Rightarrow h=\frac{2 x}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow h=x(\sqrt{3}+1)$

Hence the correct option is $a$.