Answer: 24.6 degrees
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Step-by-step explanation:
Refer to the diagram below.
If we focus on triangle ACD, then we can use the pythagorean theorem to say
d^2 + m^2 = 2050^2
d^2 = 2050^2 - m^2
Now if we focus on triangle DCB, we can also use the pythagorean theorem to say
d^2 + n^2 = 1420^2
2050^2 - m^2 + n^2 = 1420^2 ...... replace d^2 with 2050^2 - m^2
n^2 - m^2 = 1420^2 - 2050^2
n^2 - m^2 = -2,186,100
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In the diagram, I set up m and n such that m+n = 3000. This solves to n = 3000-m
We can then say
n^2 - m^2 = -2,186,100
(3000-m)^2 - m^2 = -2,186,100
m^2 - 6000m + 9,000,000 - m^2 = -2,186,100
-6000m + 9,000,000 = -2,186,100
-6000m = -2,186,100 - 9,000,000
-6000m = -11,186,100
m = (-11,186,100)/(-6000)
m = 1864.35
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We now have enough to find angle theta, which is the angle between segments AC and AD (the line of sight to the first tower and the highway lines respectively)
We'll focus entirely on triangle ACD
cos(angle) = adjacent/hypotenuse
cos(A) = AD/AC
cos(theta) = m/2050
cos(theta) = 1864.35/2050
cos(theta) = 0.90943902439024 which is approximate
theta = arccos(0.90943902439024)
theta = 24.572056388088
theta = 24.6 degrees
Make sure your calculator is in degree mode. One way to check is that cos(60) should evaluate to 0.5