The distance BC from Tower 2 to the plane will be 14,065.5 ft and the height of the plane from the ground will be 5,720.9 ft.
<h3>What is a right-angle triangle?</h3>
It is a type of triangle in which one angle is 90 degrees and it follows the Pythagoras theorem and we can use the trigonometry function.
A plane is located at C on the diagram.
There are two towers located at A and B.
The distance between the towers is 7600 feet and angles of elevation are given.
Then in the right-angle triangle ΔADC, we have
...1
Then in the right-angle triangle ΔBDC, we have
...2
From equations 1 and 2, we have
0.4452X = 0.2756 X + 2179.2649
0.1696X = 2179.2649
X = 12849.439 ≅ 12,849.4 ft
Then the distance BC from Tower 2 to the plane will be
![\rm BC = \dfrac{12849.4}{\cos 24}\\\\\\BC = 14065.5 \ ft](https://tex.z-dn.net/?f=%5Crm%20BC%20%3D%20%5Cdfrac%7B12849.4%7D%7B%5Ccos%2024%7D%5C%5C%5C%5C%5C%5CBC%20%3D%2014065.5%20%5C%20ft)
Then the height of the plane from the ground will be
![\rm H = 12849.4 \times \tan 24 \\\\\\H = 5720.9 \ ft](https://tex.z-dn.net/?f=%5Crm%20H%20%3D%2012849.4%20%5Ctimes%20%20%5Ctan%2024%20%5C%5C%5C%5C%5C%5CH%20%3D%205720.9%20%5C%20ft)
The figure is shown below.
More about the right-angle triangle link is given below.
brainly.com/question/3770177
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