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?

Answer 1

112.2ft

Explanation:

As can be seen in the attached diagram, The top of the building is at C.

The distance from the man to the building is |BD| and the height of the man is |AB| where his eye level is at B.

We are required to find the height |CE| of the building.

Now |AB|=|DE|, (opposite sides of a rectangle).

Tan 65° = |CD|/59

|CD|=59 XTan65°=107.23

The height of the building |CE|=|CD|+|DE|=107.23+5=112.23ft

The height=112.2ft ( to the nearest tenth)