A man standing on top of a building sees an automobile on a street below him. The angle of depression is 36°22'. If his eye is 550ft above the level of the automobile,how far is the automobile from him
Show Work
2006-12-16 08:15:29 · 3 answers · asked by GHz 1 in Science & Mathematics ➔ Mathematics
let
550 = height of the building
36° 22' = 36.366666667°. .Angle of Depression
cosB = will be used, since the man is on top of the building
cos 36.366666667° = 0.805238898
CosB = a/h
cos 0. 805238898 = 550/h
h(cos0.805238898) = H(550/h)
h(cos0.805238898 = 550
h(cos0.805238898/cos0.805238898= 550/0.805238898
h = 550/0.805238898
h = 683.027113 feet
The Hypotenuse is 683 feet
- - - - - - - - - -.s.
2006-12-16 10:53:36 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
You draw a right angle triangle. The opposite side is 500ft, and the angle is 36°22'.
Sine = Opposite / Hypothenuse. The hypothenuse is the distance from the man to the car. So,
sin 36°22' = 550 / Hyp
So, Hyp = 550 / sin 36°22'
Hyp = 550 / sin 36.36667
Hyp = 550 / 0.597
Hyp = 921.03 feet
2006-12-16 16:29:53 · answer #2 · answered by Renaud 3 · 0⤊ 0⤋
550 would be the opposite leg and the distance between him and the auto would be the hypotenuse so
sin(36°22') = 550/d
d = 550 / sin 36°22'
2006-12-16 16:31:22 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋