A straight vertical pole was broken during a cyclone in such a way that its top touched the ground at a distance of $6\sqrt{3}$ m from the bottom of the pole and made an angle of 30° with the horizontal. What was the height (in m) of the pole?

18

⇒ AD = AC (Broken top)

⇒ In triangle ABC

⇒ tan\({30}^\circ\) = \(\frac{AB}{BC}\)

⇒ \(\frac{1}{√3}\) = \(\frac{AB}{6√3 }\)

⇒ AB = 6m

Now,

⇒ cos\({30}^\circ\) = \(\frac{BC}{AC}\)

⇒ \(\frac{√3}{2}\) = \(\frac{6√3}{AC}\)

⇒ AC = 12m

Therefore height of the pole = AB + AC = 6 + 12 = 18m.