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Trava [24]
3 years ago
5

Click image to see it all:))))​

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0

9514 1404 393

Answer:

  16 miles

Step-by-step explanation:

The problem can be modeled by a right triangle with one angle of 7° and the side opposite being 10,000 ft. The distance needed is the hypotenuse of the triangle, so the relevant trig relation is ...

  Sin = Opposite/Hypotenuse

  Hypotenuse = Opposite/Sin

  air distance = (10,000 ft)/sin(7°) ≈ 82,055 ft

At 5,280 ft per mile, that is ...

  (82,055 ft)/(5,280 ft/mi) ≈ 15.54 mi

The plane's air distance to the airport is about 16 miles.

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URGENT! PLEASE HELP.
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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

\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

#SPJ1

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Answer:

median

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Answer:

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Step-by-step explanation:

-5x/5 - 20/5 - 4x/4 + 28/4 ➡-x - 4 -x +7 ➡-2x + 3

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