Two airplanes leave the airport. Plane A departs at a 41° angle from the runway, and plane B departs at a 43° from the runway. Which plane was farther away from the airport when it was 5 miles from the ground? Round the solutions to the nearest hundredth. Plane A because it was 7.62 miles away Plane A because it was 6.63 miles away Plane B because it was 6.84 miles away Plane B because it was 7.33 miles away

Plane A because it was 7.62 miles away.

Step-by-step explanation:

In the case of plane A we have a right triangle with an angle of 41 degrees and the opposite side  = 5.

The distance of the planes from the airport is the hypotenuse of the triangles.

sin 41 = opposite / hypotenuse

sin 41 = 5 / h       where h is the distance of the plane from the airport so

y = 5 / sin 41 = 7.62 miles.

By a similar process for Plane B  we see that sin 43 = 5/h

h = 5 / sin 43 = 7.33 miles.

answer: 19 is   2

step-by-step explanation: i just looked at your answer and wrote it down

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step-by-step explanation: