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?
Answer & explanation
Correct answer: option 1
⇒ 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.