Plane A takes off at a \( 13^{\circ} \) angle from the runway, and plane B takes off at a \( 9^{\circ} \) angle from the runway. Which plane reaches a greater horizontal distance from the airport when the plane reaches an altitude of 10,000 feet? Round the solutions to the nearest whole number. Olane A because it was 43,315 feet away Plane A because it was 63,925 feet away Plane B because it was 44,454 feet away O Plane B because it was 63,138 feet away

To determine which plane reaches a greater horizontal distance at an altitude of 10,000 feet, we can use the formula for the range of a plane taking off at an angle. The horizontal distance \( d \) can be calculated using the formula \( d = h \cdot \tan(\theta) \), where \( h \) is the altitude and \( \theta \) is the angle of elevation. For Plane A at \( 13^{\circ} \): \[ d_A = 10,000 \cdot \tan(13^{\circ}) \approx 10,000 \cdot 0.2309 \approx 2,309 \text{ feet} \] For Plane B at \( 9^{\circ} \): \[ d_B = 10,000 \cdot \tan(9^{\circ}) \approx 10,000 \cdot 0.1584 \approx 1,584 \text{ feet} \] Now, converting these distances from vertical to horizontal (using the sine function to resolve the altitude relative to the angle would yield a more accurate horizontal component). However, it’s apparent Plane A, taking off at a steeper angle, will cover a greater horizontal distance due to the larger tangent value, putting it as the winner in this takeoff match! So keep in mind, for future plane analyses, the steeper takeoff angles generally allow aircraft to climb and cover horizontal ground more effectively, making a significant difference, especially at low altitudes!