Question

nolan 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 7325 feet. nolan initially measures an angle of elevation of 15° to the plane at point a. at some later time, he measures an angle of elevation of 36° to the plane at point b. find the distance the plane traveled from point a to point b. round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Calculate the horizontal distance from point \(A\)

We use the tangent function \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
For the angle of elevation of \(15^{\circ}\), let the horizontal distance from the initial - position (point \(A\)) to the point directly below the plane be \(x_1\).
Since \(\tan15^{\circ}=\frac{7325}{x_1}\), then \(x_1 = \frac{7325}{\tan15^{\circ}}\).
We know that \(\tan15^{\circ}=2 - \sqrt{3}\approx0.2679\), so \(x_1=\frac{7325}{0.2679}\approx27342.29\)

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

For the angle of elevation of \(36^{\circ}\), let the horizontal distance from the later - position (point \(B\)) to the point directly below the plane be \(x_2\).
Since \(\tan36^{\circ}=\frac{7325}{x_2}\), then \(x_2=\frac{7325}{\tan36^{\circ}}\).
We know that \(\tan36^{\circ}\approx0.7265\), so \(x_2=\frac{7325}{0.7265}\approx10082.6\)

Step3: Calculate the distance \(AB\)

The distance the plane traveled from point \(A\) to point \(B\) is \(d=x_1 - x_2\)
\(d = 27342.29-10082.6=17259.7\)

Answer:

\(17259.7\) feet