A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m.

Let AB be the tower and BC be the flagstaff.

We have,

$\mathrm{BC}=6 \mathrm{~m}, \angle \mathrm{AOB}=30^{\circ}$ and $\angle \mathrm{AOC}=60^{\circ}$

Let $\mathrm{AB}=h$

In $\triangle \mathrm{AOB}$,

$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{OA}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{OA}}$

$\Rightarrow \mathrm{OA}=h \sqrt{3} \quad \ldots$ (i)

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

$\tan 60^{\circ}=\frac{\mathrm{AC}}{\mathrm{OA}}$

$\Rightarrow \sqrt{3}=\frac{\mathrm{AB}+\mathrm{BC}}{h \sqrt{3}} \quad[\mathrm{U} \operatorname{sing}(\mathrm{i})]$

$\Rightarrow 3 h=h+6$

$\Rightarrow 3 h-h=6$

$\Rightarrow 2 h=6$

$\Rightarrow h=\frac{6}{2}$

$\Rightarrow h=3 \mathrm{~m}$

So, the height of the tower is 3 m.