Question

fatoumata 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. fatoumata initially measures an angle of elevation of 17° to the plane at point a. at some later time, she 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: Calculate the horizontal distance from point \(A\)

We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For the angle of elevation at point \(A\) (\(\theta = 17^{\circ}\)) and the altitude (opposite side) \(h = 6875\) feet. Let \(x_{A}\) be the horizontal distance from point \(A\) to the point directly below the plane. Then \(\tan(17^{\circ})=\frac{6875}{x_{A}}\), so \(x_{A}=\frac{6875}{\tan(17^{\circ})}\). Using a calculator, \(\tan(17^{\circ})\approx0.3057\), and \(x_{A}=\frac{6875}{0.3057}\approx22500\) feet.

Step2: Calculate the horizontal distance from point \(B\)

For the angle of elevation at point \(B\) (\(\theta = 40^{\circ}\)) and the altitude \(h = 6875\) feet. Let \(x_{B}\) be the horizontal distance from point \(B\) to the point directly below the plane. Then \(\tan(40^{\circ})=\frac{6875}{x_{B}}\), so \(x_{B}=\frac{6875}{\tan(40^{\circ})}\). Using a calculator, \(\tan(40^{\circ})\approx0.8391\), and \(x_{B}=\frac{6875}{0.8391}\approx8200\) feet.

Step3: Calculate the distance between \(A\) and \(B\)

The distance \(d\) that the plane traveled from \(A\) to \(B\) is \(d=x_{A}-x_{B}\). Substitute the values of \(x_{A}\) and \(x_{B}\): \(d = 22500 - 8200=14300\) feet.

Answer:

\(14300\) feet