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 / 7.62 mi away

Plane A / 6.63 mi away

Plane B / 6.84 mi away

Plane B / 7.33 mi away

Plane A / 7.62 mi 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.

Therefore Plane A was farther away from the airport (7.62 miles away) when it was 5 miles from the ground.

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step-by-step explanation: