Question

URGENT! PLEASE HELP.

Answer 1

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.

What is a right-angle triangle?

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

$\rm \tan 16 = \dfrac{H}{7600 + X}\\\\\\H = 0.2867(7600 +X)\\\\\\H = 0.2756\ X + 2179.2649$ ...1

Then in the right-angle triangle ΔBDC, we have

$\rm \tan24 = \dfrac{H}{ X}\\\\\\H = 0.4452\ X$ ...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

$\rm BC = \dfrac{12849.4}{\cos 24}\\\\\\BC = 14065.5 \ ft$

Then the height of the plane from the ground will be

$\rm H = 12849.4 \times \tan 24 \\\\\\H = 5720.9 \ ft$

The figure is shown below.

More about the right-angle triangle link is given below.

brainly.com/question/3770177

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