Question

savannah 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 5450 feet. savannah initially measures an angle of elevation of 15° to the plane at point a. at some later time, she measures an angle of elevation of 29° 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.

Explanation:

Step1: Find horizontal distance to A

Let $x_A$ = horizontal distance from Savannah to point directly below A. Use $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$, so:
$$x_A = \frac{5450}{\tan(15^\circ)}$$
$\tan(15^\circ) \approx 0.2679$, so $x_A \approx \frac{5450}{0.2679} \approx 20343.41$ feet.

Step2: Find horizontal distance to B

Let $x_B$ = horizontal distance from Savannah to point directly below B. Use the same tangent formula:
$$x_B = \frac{5450}{\tan(29^\circ)}$$
$\tan(29^\circ) \approx 0.5543$, so $x_B \approx \frac{5450}{0.5543} \approx 9832.22$ feet.

Step3: Calculate distance AB

The plane travels horizontally between A and B, so $AB = x_A - x_B$:
$$AB \approx 20343.41 - 9832.22 = 10511.19$$

Answer:

10511 feet