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: Recall tangent function

In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side to the angle of elevation is the altitude (\(h = 6875\) feet), and the adjacent side is the horizontal distance from Fatoumata to the point directly below the plane. Let \(x_A\) be the horizontal distance at point \(A\) and \(x_B\) at point \(B\). So, \(\tan(17^\circ)=\frac{6875}{x_A}\) and \(\tan(40^\circ)=\frac{6875}{x_B}\).

Step2: Solve for \(x_A\) and \(x_B\)

For \(x_A\): \(x_A=\frac{6875}{\tan(17^\circ)}\). Calculate \(\tan(17^\circ)\approx0.3057\), so \(x_A\approx\frac{6875}{0.3057}\approx22489.37\) feet.
For \(x_B\): \(x_B=\frac{6875}{\tan(40^\circ)}\). Calculate \(\tan(40^\circ)\approx0.8391\), so \(x_B\approx\frac{6875}{0.8391}\approx8193.30\) feet.

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

The distance the plane travels (\(d\)) is \(x_A - x_B\) (since it's approaching, \(x_A>x_B\)). So, \(d = 22489.37 - 8193.30=14296.07\approx14296\) feet.

Answer:

The distance the plane traveled from point \(A\) to point \(B\) is \(\boxed{14296}\) feet.