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?
Answer & explanation
Correct answer: option 3
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