Question

kayla spots an airplane on radar that is currently approaching in a straight line, and that will fly directly overhead. the plane maintains a constant altitude of 6875 feet. kayla initially measures an angle of elevation of 16° to the plane at point a. at some later time, she measures an angle of elevation of 30° to the plane at point b. find the distance the plane traveled from point a to point b. round your answer to the nearest foot if necessary. answer attempt 2 out of 2

Explanation:

Step1: Find horizontal distance for point A

Let $x_A$ = horizontal distance from Kayla to point A. Use $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$, so:
$$x_A = \frac{6875}{\tan(16^\circ)}$$
Calculate $\tan(16^\circ) \approx 0.2867$, so $x_A \approx \frac{6875}{0.2867} \approx 24009$ feet.

Step2: Find horizontal distance for point B

Let $x_B$ = horizontal distance from Kayla to point B. Use the same tangent formula:
$$x_B = \frac{6875}{\tan(30^\circ)}$$
$\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.5774$, so $x_B \approx \frac{6875}{0.5774} \approx 11907$ feet.

Step3: Calculate distance from A to B

The plane's path is straight, so distance $AB = x_A - x_B$ (since it is approaching):
$$AB \approx 24009 - 11907 = 12102$$

Answer:

12102 feet