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? |
30.46 m 37.34 m 36.84 m 34.64 |
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 |