1. A tourist sights the Hancock tower from the highway with an angle of elevation of 12°. If the

Hancock tower is known to be 1350 feet high, how far away is the tourist, to the nearest tenth?

Respuesta :

Answer: [tex]6351.25\ ft[/tex]

Step-by-step explanation:

Given

Angle of elevation [tex]x=12^{\circ}[/tex]

Height of tower [tex]h=1350\ ft[/tex]

Suppose tourist is d feet away from the tower

from the figure, we can write

[tex]\Rightarrow \tan 12^{\circ}=\dfrac{1350}{d}\\\\\Rightarrow d=\dfrac{1350}{\tan 12^{\circ}}=6351.25\ ft[/tex]

Thus, the distance of tourist from the tower is [tex]6350.25\ ft[/tex]

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