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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

Respuesta :

sqdancefan

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Answer:

  (b)  112.2 ft

Step-by-step 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

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