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

Answer:

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

Step-by-step 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

$tan(30\°)=\frac{h}{50}$

$h=tan(30\°)(50)=28.87\ ft$

Find the height of the building H

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

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