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

in progress 0
Linh Đan 2 months 2021-07-22T19:22:38+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-07-22T19:24:34+00:00

    9514 1404 393

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )