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 le
vel of 5 ft. How tall is the building to the nearest tenth of a foot?
1 answer:
Answer:
112.2ft
Step-by-step 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)
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