Question

A student stands 50 feet away from the front of a building and measures the angle of elevation to the top of the building. From the student's eye level 5 feet off the ground, the angle of elevation to the top of the building is 30^\circ. Approximately, what is the height of the building?

Answer 1

The height of the building is approximately
`34\ ft`

Explanation:

Let

h------> height of the building from the student's eye level

H----> height of the building from the ground (H=h+5 ft)

we know that

`tan(30\°)=(h)/(50)`

`h=tan(30\°)(50)=28.87\ ft`

Find the height of the building H

`H=h+5=28.87+5=33.87\ ft`

The height of the building is approximately
`34\ ft`