Question

question 7 (multiple choice worth 1 points) 05.03r mc) plane a takes off at a 13° angle from the runway, and plane b takes off at a 9° 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. plane 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 plane b because it was 63,138 feet away

Explanation:

Step1: Use the tangent function

We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side is the altitude (\(h = 10000\) feet) and the adjacent side is the horizontal distance (\(x\)). So, \(x=\frac{h}{\tan\theta}\)

Step2: Calculate for Plane A

For Plane A, \(\theta = 13^{\circ}\), \(h = 10000\) feet. Then \(x_A=\frac{10000}{\tan(13^{\circ})}\). Since \(\tan(13^{\circ})\approx0.2309\), \(x_A=\frac{10000}{0.2309}\approx43315\) feet

Step3: Calculate for Plane B

For Plane B, \(\theta = 9^{\circ}\), \(h = 10000\) feet. Then \(x_B=\frac{10000}{\tan(9^{\circ})}\). Since \(\tan(9^{\circ})\approx0.1584\), \(x_B=\frac{10000}{0.1584}\approx63138\) feet

Answer:

Plane B because it was 63,138 feet away