Description for Correct answer:
Let the height of pole AD = BC = h



In \( \Large \triangle OBC \tan 60 ^{\circ} = \frac{h}{x} \)

=> \( \Large x = \frac{h}{\sqrt{3}} \)

In \( \Large \triangle AOD \tan 30 ^{\circ} = \frac{h}{100-x} \)

=> \( \Large h = \left(100 - x\right) \frac{1}{\sqrt{3}}

= \left(100 - \frac{h}{\sqrt{3}}\right) \frac{1}{\sqrt{3}} \) [from (i)]

=> \( \Large 3h = 100\sqrt{3} - h \)

=> \( \Large h = \frac{100\sqrt{3}}{4} \)

=> \( \Large h = 25\sqrt{3} m \)