Question

question 6 (multiple - choice worth 1 point) plane a takes off at a 8° angle from the runway, and plane b takes off at a 11° 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 51,446 feet away. plane a because it was 71,154 feet away. plane b because it was 52,408 feet away. plane b because it was 71,853 feet away.

Explanation:

Step1: Use tangent - function for plane A

We know that $\tan\theta=\frac{opposite}{adjacent}$. For plane A, $\theta = 8^{\circ}$ and the opposite side (altitude) $h = 10000$ feet. Let the horizontal distance be $x_A$. Then $\tan(8^{\circ})=\frac{10000}{x_A}$, so $x_A=\frac{10000}{\tan(8^{\circ})}$. Since $\tan(8^{\circ})\approx0.1405$, $x_A=\frac{10000}{0.1405}\approx71153$ feet.

Step2: Use tangent - function for plane B

For plane B, $\theta = 11^{\circ}$ and the opposite side (altitude) $h = 10000$ feet. Let the horizontal distance be $x_B$. Then $\tan(11^{\circ})=\frac{10000}{x_B}$, so $x_B=\frac{10000}{\tan(11^{\circ})}$. Since $\tan(11^{\circ})\approx0.1944$, $x_B=\frac{10000}{0.1944}\approx51440$ feet.

Step3: Compare the horizontal distances

Since $71153>51440$, plane A reaches a greater horizontal distance.

Answer:

Plane A because it was 71154 feet away (rounded to the nearest whole number)