Question

question
corey 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 6800 feet. corey initially measures an angle of elevation of 16° to the plane at point a. at some later time, he measures an angle of elevation of 40° 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 the horizontal distance when angle is \(40^{\circ}\)

Let the horizontal distance from the point directly below the plane to the observer be \(x_1\) when the angle of elevation is \(40^{\circ}\). We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Given the altitude (opposite side) \(h = 6800\) feet and \(\theta = 40^{\circ}\), then \(\tan40^{\circ}=\frac{6800}{x_1}\), so \(x_1=\frac{6800}{\tan40^{\circ}}\). Using a calculator, \(\tan40^{\circ}\approx0.8391\), then \(x_1=\frac{6800}{0.8391}\approx8104\) feet.

Step2: Find the horizontal distance when angle is \(16^{\circ}\)

Let the horizontal distance from the point directly below the plane to the observer be \(x_2\) when the angle of elevation is \(16^{\circ}\). Using \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), with \(h = 6800\) feet and \(\theta = 16^{\circ}\), then \(\tan16^{\circ}=\frac{6800}{x_2}\), so \(x_2=\frac{6800}{\tan16^{\circ}}\). Using a calculator, \(\tan16^{\circ}\approx0.2867\), then \(x_2=\frac{6800}{0.2867}\approx23718\) feet.

Step3: Calculate the distance the plane traveled

The distance the plane traveled \(d=x_2 - x_1\). Substitute \(x_1\approx8104\) and \(x_2\approx23718\) into the formula: \(d = 23718-8104=15614\) feet.

Answer:

\(15614\) feet