Correct option is: A) 10 √3 m
In right \(\triangle\)ABC
\(\angle\)CAB = \(30^\circ\) & AB = 30 m
tan A = \(\frac {BC}{AB}\)
= BC = AB tan A
= AB tan \(30^\circ\) (\(\because\) \(\angle\)CAB = \(30^\circ\))
= \(\frac {AB}{\sqrt3}\) (\(\because\) tan \(30^\circ\) = \(\frac 1{\sqrt3}\))
= \(\frac {30}{\sqrt3}\) (\(\because\) AB = 30 m)
= 10\(\sqrt3\) m
Hence, the height of the tower is 10\(\sqrt3\) m.