If an observer is 200 feet from a building and the angle of elevation to the top of a building is 60degrees, what is the height of the building?
2007-06-05 07:08:26 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
tanx = opposite / adjacent
tan60 = height/200
height = 346.4 ft
2007-06-05 07:11:23 · answer #1 · answered by Kemmy 6 · 1⤊ 0⤋
30-60-90 triangles are nice because their sides have a ratio of 1-2-√3. Therefore the answer is:
200√3 = 346.41016151377545870548926830117
2007-06-05 07:30:26 · answer #2 · answered by Dave 6 · 0⤊ 0⤋
My height or taller. Athletic build. Doesn't need to have a lot of muscles but needs to have some shape, not just flat as a board.
2016-05-17 10:24:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
let height = h ft
tan 60° = h / 200
h = 200.tan 60°
h = 346.4 ft
2007-06-05 07:12:30 · answer #4 · answered by Como 7 · 0⤊ 0⤋
CAN BE REPRESENTED BY:
cos(60 deg) = 200 feet / h
h = 400 feet
THEREFORE, HEIGHT IS:
sin(60 deg) = x / h
sin(60 deg) = x / 400 feet
x = 200 (SQRT(3))
CHECK OUR SOLUTION:
tan(60 deg) = x / 200 feet
x = 200 times (SQRT(3))
THEREFORE, HEIGHT IS: 200 times (SQRT(3))
2007-06-05 07:31:58 · answer #5 · answered by Rey Arson II 3 · 0⤊ 0⤋