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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

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

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Mathematics
2 months
2021-07-22T19:22:38+00:00
2021-07-22T19:22:38+00:00 1 Answers
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## Answers ( )

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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