A landscaper is planting a row of 10 shrubs along the walkway shown below. There must be one shrub at the very beginning and one
shrub at the very end, and the shrubs in between will be equally spaced along the length of the walkway. Which equation can the landscaper use to find the distance d, in feet, to leave between the shrubs?
1 answer:
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Answer:
a. distance of the surveyor to the base of the building = 2051.90 ft
b. height of the building = 1384 ft
c. Angle of elevation from the surveyor to the top of the antenna = 38.31°
d. Height of antenna = 237.08 ft
Step-by-step explanation:
The picture above is a illustration of the described event.
a = the height of the flag
b = the height of the building
c = distance of the surveyor from the base of the building
the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°
distance from her position to the top of the building = 2475 ft
distance from her position to the top of the flag = 2615 ft
(a) How far away from the base of the building is the surveyor located?
using the SOHCAHTOA principle
cos 34° = c/2475
c = 0.8290375726 × 2475
c = 2051.8679921
c = 2051.90 ft
(b) How tall is the building
The height of the building = b
sin 34° = opposite /hypotenuse
0.5591929035 = b/2475
b = 0.5591929035 × 2475
b = 1384.0024361
b = 1384.00 ft
(c) What is the angle of elevation from the surveyor to the top of the antenna?
let the angle = ∅
cos ∅ = adjacent/hypotenuse
cos ∅ = 2051.90/2615
cos ∅ = 0.784665392
∅ = cos-1 0.784665392
∅ = 38.310258303
∅ = 38.31°
(d) How tall is the antenna?
height of the antenna = a
sin 38.31° = opposite/hypotenuse
sin 38.31° = (a + b)/2615
sin 38.31° × 2615 = (a + b)
(a + b) = 0.6199159917 × 2615
(a + b) = 1621.0803182
(a + b) = 1621. 08 ft
Height of antenna = 1621. 08 - 1384.00 = 237.08031822 ft
Height of antenna = 237.08 ft
