two airplanes leave the airport. plane a departs at a 44° angle from the runway, and plane b departs at a 40° from the runway. which plane was farther away from the airport when it was 22 miles from the ground? round the solutions to the nearest hundredth.
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

Plane B was farther away from the airport when it was 22 miles from the ground because Plane B was 34.23 miles away

Step-by-step explanation:

The journey of both planes can be represented by two right angled triangles and the distance of the planes from the airport can be represented by the hypotenuse

Solution for plane A:

We have angle and side opposite to angle so we will use sine to calculate the hypotenuse:

sinθ= Opposite/hypotenuse

hypotenuse = opposite/sinθ

hypotenuse = 22/sin(44)

hypotenuse = 31.67

Plane A is 31.67 miles away from the airport

Plane B:

sinθ= Opposite/hypotenuse

hypotenuse = opposite/sinθ

hypotenuse = 22/sin(40)

hypotenuse = 34.23

It means that Plane B is 34.23 miles away from airport.

Therefore it is concluded that Plane B was farther away from the airport when it was 22 miles from the ground because Plane B was 34.23 miles away

283 square feet

step-by-step explanation:

good luck