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:

To solve this question, we would be using Trigonometric function for Sine

For plane A

Sin θ = Opposite side / Hypotenuse

For Plane A, θ =41 °

Opposite side = Height = 5 miles

Hypotenuse = Distance = Unknown

sin(41°)= 5 / Hypotenuse

Hypotenuse = 5 / sin 41°

=7.6212654335 miles

Approximately = 7.62 miles.

For plane B

Sin θ = Opposite side / Hypotenuse

For Plane B, θ =43 °

Opposite side = Height = 5 miles

Hypotenuse = Distance = Unknown

sin(43°)= 5 / Hypotenuse

Hypotenuse = 5 / sin 41°

=7.3313959282 miles

Approximately = 7.33 miles.

Therefore by comparing the distance of both planes as calculated above, the plane that was farther away from the airport when it was 5 miles from the ground is Plane A because it was 7.62 miles away.

idk bring the function back   im not smart

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