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:

27.6

⇒ 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.