John wants to invite around 78,000 college students to a tournament. Which of the following sets of college student populations
2 answers:
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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
