Let the height of the tower be h.



\( \Large In \triangle BCD, \tan 45 ^{\circ} = \frac{h}{BC} \)

=> BC = h

\( \Large In \triangle ABD, \tan 30 ^{\circ} = \frac{h}{AB} \)

=> \( \Large AB = \frac{h}{\tan 30 ^{\circ} } \)

=> \( \Large AB= \sqrt{3}h => AC+BC=\sqrt{3}h \)

=> \( \Large 60 + h = \sqrt{3}h \) [using (i)]

=> \( \Large h = \frac{60}{\sqrt{3}-1} = \frac{60 \left(\sqrt{3}+1\right) }{2} = 30 \left(\sqrt{3}+1\right) m \)