An observer who is 1.62 m tall is 45 m away from a pole. The angle of elevation of the top of the pole from his eyes is $30^\circ$. The height (in m) of the pole is closest to:
Answer & explanation
Correct answer: option 3
⇒ tan\({30}^\circ\) = \(\frac{AB}{45}\) (BE = CD)
⇒ \(\frac{1}{√3}\) = \(\frac{AB}{45}\)
⇒ AB = \(\frac{45}{√3}\) = 15\(\sqrt {3 }\)m
⇒ AC = AB + BC
⇒ AC = 15\(\sqrt {3 }\) + 1.62 = 27.6m.
Therefore, height of the pole is 27.6m.