Choose the correct answer of the following question:

Solution:

Let CD = h be the height of the tower.

We have,

$\mathrm{AB}=2 x, \angle \mathrm{DAC}=30^{\circ}$ and $\angle \mathrm{DBC}=45^{\circ}$

In $\Delta \mathrm{BCD}$,

$\tan 45^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}}$

$\Rightarrow 1=\frac{h}{\mathrm{BC}}$

$\Rightarrow \mathrm{BC}=h$

Now, in $\triangle \mathrm{ACD}$,

$\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AB}+\mathrm{BC}}$

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

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

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

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

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

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

$\Rightarrow h=\frac{2 x(\sqrt{3}+1)}{2}$

$\therefore h=x(\sqrt{3}+1) \mathrm{m}$

Hence, the correct answer is option (d).