One plane flies straight east at an altitude of 31,000 feet. A second plane is flying west at an altitude of 13,000 feet on a course that lies directly below that of the first plane and directly above the straight road from Thomasville to Johnsburg. As the first plane passes over Thomasville, the second is passing over Johnsburg. At that instant both planes spot a beacon next to the road between Thomasville to Johnsburg. The angle of depression from the first plane to the beacon is 62°, and the angle of depression from the second plane to the beacon is 34°. How far is Thomasville from Johnsburg?

cot62 - cot 34

To find the distance between Thomasville and Johnsburg, we need to utilize the concept of trigonometry. Let's denote the distance between Thomasville and Johnsburg as 'x' (in units).

Firstly, let's consider the first plane. The angle of depression from the first plane to the beacon is 62°. This angle corresponds to the angle between the line of sight from the plane to the beacon and the horizontal plane. Since the plane is flying straight east and the beacon is straight down from the plane, we can consider this angle as the angle of elevation from the beacon to the plane. We can use the tangent function to find the relationship between the angle of elevation and the opposite side (altitude of the plane) relative to the adjacent side (distance from the beacon to Thomasville).

Using the tangent function:

tan(62°) = opposite (altitude of the plane) / adjacent (distance between the beacon and Thomasville)
tan(62°) = 31,000 / x

Similarly, for the second plane, the angle of depression from the second plane to the beacon is 34°. Again, this angle corresponds to the angle of elevation from the beacon to the plane. Using the same reasoning, we can find the relationship between the angle of elevation and the opposite side (altitude of the plane) relative to the adjacent side (distance from the beacon to Johnsburg).

Using the tangent function:

tan(34°) = opposite (altitude of the plane) / adjacent (distance between the beacon and Johnsburg)
tan(34°) = 13,000 / x

Now, we have two equations with two variables. We can solve this system of equations to find the value of 'x', which represents the distance between Thomasville and Johnsburg.

Using the given angle measurements and the equations derived:

tan(62°) = 31,000 / x
tan(34°) = 13,000 / x

We can rearrange the first equation to solve for x:

x = 31,000 / tan(62°)
x = 31,000 / 2.1445
x ≈ 14,461.56

Similarly, rearranging the second equation:

x = 13,000 / tan(34°)
x = 13,000 / 0.6641
x ≈ 19,568.77

Since we are looking for the distance between Thomasville and Johnsburg, we can take the average of these two values:

Average distance = (14,461.56 + 19,568.77) / 2
Average distance ≈ 17,015.17

Therefore, the distance between Thomasville and Johnsburg is approximately 17,015.17 units.