Correct option is (C) 30√3 मी०
Let the height of tower = h m
In \(\triangle A B C\)
\(\tan 60^{\circ} =\frac{A B}{B C}\)
\(\sqrt{3} =\frac{h}{30 \mathrm{~m}}\quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]\)
\(h =30 \sqrt{3} \mathrm{~m}\)
Therefore, the height of the vertical tower is 30√3 m.
