Question

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?

Answer 1

Answer:
`6351.25\ ft`

Explanation:

Given

Angle of elevation
`x=12^(\circ)`

Height of tower
`h=1350\ ft`

Suppose tourist is d feet away from the tower

from the figure, we can write

`\Rightarrow \tan 12^(\circ)=(1350)/(d)\\\\\Rightarrow d=(1350)/(\tan 12^(\circ))=6351.25\ ft`

Thus, the distance of tourist from the tower is
`6350.25\ ft`