from a window 20 feet above the ground, the angle of elevation to the top of another building is 35 degrees. The distance betwee
n the building is 52 feet. Find the height of the building nearest to the nearest tenth of a foot
1 answer:
Answer:
56.4 ft
Step-by-step explanation:
The tangent of the angle relates the opposite side (height above the window) to the adjacent side (distance between buildings) in the right triangle that models this geometry.
Tan = Opposite/Adjacent
tan(35°) = (height above window)/(distance between buildings)
height above window = (52 ft)tan(35°) ≈ 36.4 ft
Added to the window height of 20 ft, the height of the other building is ...
20 ft + 36.4 ft = 56.4 ft
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Quadrant IV = only cosine is positive
a = height (4)
b = base ( 8^2-4^2=b^2
b = 6.93 @
c = hypothenuse(8)
cos =
/8
tan = - 4/
sec = 1/cos
1/cos = 1/ (
/8)
sec = 8/
csc = 1/sin
1/sin = 1/(-4/8)
csc = - 2
cot = 1/tan
1/tan = 1/(-4/
)
cot = -
/4
Quadrant IV = only cosine is positive
a = height (4)
b = base ( 8^2-4^2=b^2
b = 6.93 @
c = hypothenuse(8)
cos =
/8
tan = - 4/
sec = 1/cos
1/cos = 1/ (
/8)
sec = 8/
csc = 1/sin
1/sin = 1/(-4/8)
csc = - 2
cot = 1/tan
1/tan = 1/(-4/
)
cot = -
/4