The height of a tower is 100 m. When the angle of elevation of the sun

Solution:

The given situation can be represented as,

Here, AB is the tower of height  meters.

When angle of elevation of sun changes from $\angle D=30^{\circ}$ to $\angle C=45^{\circ}, C D=x$.

We assumed that $B C=y$

Here we have to find the value of $x$

So we use trigonometric ratios.

In a triangle,

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

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

$\Rightarrow \mathrm{I}=\frac{100}{y}$

$\Rightarrow y=100$

Again in a triangle ABD,

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

$\Rightarrow \tan 30^{\circ}=\frac{100}{x+y}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{x+y}$

$\Rightarrow 100 \sqrt{3}=x+y$

$\Rightarrow 100 \sqrt{3}=x+100$

$\Rightarrow x=100(\sqrt{3}-1)^{\mathrm{P}}$   Put $x=100$

Hence the correct option is c.