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?
Answer & explanation
Correct answer: option 2
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