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Two airplanes leave the airport. Plane A departs at a 42° angle from the runway, and plane B departs at a 44° angle from the runway. Which plane was farther away from the airport when it was 7 miles from the ground? Round the solutions to the nearest hundredth. Plane A because it was 9. 428 miles away Plane A because it was 10. 46 miles away Plane B because it was 10. 08 miles away Plane B because it was 9. 73 miles away.

Respuesta :

Using the slope concept, it is found that Plane A was farther away.

What is a slope?

  • The slope is given by the vertical change divided by the horizontal change.
  • It's also the tangent of the angle of depression.

For Plane A:

  • The vertical change is of 7 miles.
  • The horizontal change is of [tex]x_A[/tex], which is the distance we want to find.
  • The angle is of 42º.

Hence:

[tex]\tan{42^{\circ}} = \frac{7}{x_A}[/tex]

[tex]x_A = \frac{7}{\tan{42^{\circ}}}[/tex]

[tex]x_A = 7.77[/tex]

For Plane B, we have that:

  • The vertical change is of 7 miles.
  • The horizontal change is of [tex]x_B[/tex], which is the distance we want to find.
  • The angle is of 44º.

Hence:

[tex]\tan{44^{\circ}} = \frac{7}{x_B}[/tex]

[tex]x_B = \frac{7}{\tan{44^{\circ}}}[/tex]

[tex]x_B = 7.25[/tex]

7.77 > 7.25, hence Plane A was farther away.

You can learn more about the slope concept at https://brainly.com/question/26342863

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