Question

Two airplanes leave the airport. Plane A departs at a 44° angle from the runway, and plane B departs at a 40° angle from the runway. Which plane was farther away from the airport when it was 6 miles from the ground? Round the solutions to the nearest hundredth.
Group of answer choices

Plane A because it was 31.67 miles away

Plane A because it was 22.78 miles away

Plane B because it was 34.23 miles away

Plane B because it was 26.22 miles away

Answer 1

Answer: Answer:

Plane B was farthest away from the airport

Step-by-step explanation:

"This question requires you to visualize the run way as the horizontal distance to be covered, the height from the ground as the height gained by the plane after take of and the distance from the airport as the displacement due to the angle of take off.

In plane A

The take-off angle is 44° and the height gained is 22 ft.

Apply the relationship for sine of an angle;

Sine Ф°= opposite side length÷hypotenuse side length

The opposite side length is the height gained by plane which is 22 ft

The angle is 44° and the distance the plane will be away from the airport after take-off will be represented by the value of hypotenuse

Applying the formula

sin Ф=O/H where O=length of the side opposite to angle 44° and H is the hypotenuse

In plane B

Angle of take-off =40°, height of plane=22miles finding the hypotenuse

Solution

After take-off and reaching a height of 22 ft from the ground, plane A will be 31.67 miles from the airport

After take-off and reaching a height of 22 ft from the ground, plane B will be 34.23 miles away from the airport."

Excerpt from Brainly

Answer 2

Your answer is Plane B was farthest away from the airport.