On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is done by a process called triangulation, which works by using the distance from two known points. Suppose there are two cell phone towers within range of you, located 3000 feet apart along a straight highway that runs east to west, and you know you are north of the highway. Based on the signal delay, it can be determined you are 2050 feet from the first tower, and 1420 feet from the second. Determine the angle, \thetaθ, between your line of sight to the first tower and the highway to the nearest tenth of a degree. (A calculator is needed for this question)

========================================================

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

--------------------------

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

--------------------------

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

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: