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°. The height (in m) of the pole is closest to?

27.6 m

AB is observer and CD is pole

In triangle APC :

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

                (PC)          (AP)

                   ↓              ↓

                   ↓             45

             \(\frac{45}{\sqrt {3}}\)

PC = 15 \(\sqrt {3}\) = 25.98

height of pole (CD) = CP + PD = 25.98 + 1.62 = 27.6 m