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.
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
[tex]\rm \tan 16 = \dfrac{H}{7600 + X}\\\\\\H = 0.2867(7600 +X)\\\\\\H = 0.2756\ X + 2179.2649[/tex] ...1
Then in the right-angle triangle ΔBDC, we have
[tex]\rm \tan24 = \dfrac{H}{ X}\\\\\\H = 0.4452\ X[/tex] ...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
[tex]\rm BC = \dfrac{12849.4}{\cos 24}\\\\\\BC = 14065.5 \ ft[/tex]
Then the height of the plane from the ground will be
[tex]\rm H = 12849.4 \times \tan 24 \\\\\\H = 5720.9 \ ft[/tex]
The figure is shown below.
More about the right-angle triangle link is given below.
https://brainly.com/question/3770177
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