An observer who is 2.20 m tall is 60 m away from a pole. The angle of elevation of the top of the pole from his eyes is 30°. The height (in m) of the pole is closest to?

36.84 m

AB is observer and CD is pole

In triangle APC :

tan 30°  =   1     :      \(\sqrt {3}\)

                (PC)          (AP)

                   ↓              ↓

                   ↓             60

             \(\frac{60}{\sqrt {3}}\)

PC = 20 \(\sqrt {3}\) = 34.64

height of pole (CD) = CP + PD =34.64 + 2.20 = 36.84 m