Two airplanes leave the airport. Plane A departs at a 41° angle from the runway, and plane B departs at a 43° angle 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. Question 9 options:

Plane A because it was 20.71 miles away

Plane A because it was 27.43 miles away

Plane B because it was 16.79 miles away

Plane B because it was 26.39 miles away

There's no need to answer this question, bc I BLEW THE AIRPLANE UP!-

Step-by-step explanation: MWAHAHAHAH- may I have brainliest? .,.

Plane A because it was 27.43 miles away

Step-by-step explanation:

We have to calculate the distance of each plane from the airport.

In other to do this, we would use the trigonometric function of Sine

sin θ = Opposite/Hypotenuse

For Plane A

Plane A departs at a 41° angle from the runway

sin θ = Opposite/Hypotenuse

θ = 41°

Distance from the ground = Opposite = 5 miles

Hypotenuse = ???

sin 41° = 5 miles/Hypotenuse

sin 41° × Hypotenuse = 5 miles

Hypotenuse = 5 miles/sin 41°

Hypotenuse = 7.6212654335 miles

Approximately to the nearest hundredth ≈ 7.62 miles

For Plane B

Plane B departs at a 43° from the runway

sin θ = Opposite/Hypotenuse

θ = 43°

Distance from the ground = Opposite = 5 miles

Hypotenuse = ???

sin 43° = 5 miles/Hypotenuse

sin 43° × Hypotenuse = 5 miles

Hypotenuse = 5 miles/sin 43°

Hypotenuse = 7.3313959282 miles

Approximately to the nearest hundredth ≈ 7.33 miles

From the above calculation, we can see that Plane A what 7.62 miles away from the airport while Plane b was 7.33 miles away from the airport.

~Kandy~

Hope this helped!

Brainliest please!

step-by-step explanation: