Question

A person can see the top of a building at an angle of 65°. The person is standing 50 ft away from

the building and has an eye level of 5 ft. How tall is the building to the nearest tenth of a foot?
O 107.2 ft
O 112.2 ft
O 50.3 ft
O 26.1 ft

Answer 1

9514 1404 393

Answer:

(b) 112.2 ft

Explanation:

The relevant trig relation is ...

Tan = Opposite/Adjacent

For the given geometry, this becomes ...

tan(65°) = (height above eye level)/(50 ft)

Then we have ...

(height above eye level) = (50 ft)tan(65°) = 107.2 ft

Adding the height of eye level will give us the height of the building.

building height = (eye level height) + (height above eye level)

building height = (5 ft) + (107.2 ft)

building height = 112.2 ft