Question

question
keilantra 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 7500 feet. keilantra initially measures an angle of elevation of 19° to the plane at point a. at some later time, she measures an angle of elevation of 34° 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.
answer attempt 3 out of 3
feet
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Explanation:

Step1: Define the right triangles

Let's consider two right triangles. Let \( h = 7500 \) feet (the altitude of the plane, which is constant). Let \( x_1 \) be the horizontal distance from Keilantra to the point directly below the plane at point \( A \), and \( x_2 \) be the horizontal distance from Keilantra to the point directly below the plane at point \( B \). The distance the plane travels from \( A \) to \( B \) is \( d=x_1 - x_2 \) (since the plane is approaching, \( x_1>x_2 \)).

For the angle of elevation \( \theta_1 = 19^\circ \) at point \( A \), we use the tangent function: \( \tan(\theta_1)=\frac{h}{x_1} \), so \( x_1=\frac{h}{\tan(\theta_1)} \).

For the angle of elevation \( \theta_2 = 34^\circ \) at point \( B \), we use the tangent function: \( \tan(\theta_2)=\frac{h}{x_2} \), so \( x_2=\frac{h}{\tan(\theta_2)} \).

Step2: Calculate \( x_1 \)

We know \( h = 7500 \) and \( \theta_1=19^\circ \). Then \( x_1=\frac{7500}{\tan(19^\circ)} \). Since \( \tan(19^\circ)\approx0.3443 \), we have \( x_1=\frac{7500}{0.3443}\approx21783.33 \) feet.

Step3: Calculate \( x_2 \)

We know \( h = 7500 \) and \( \theta_2 = 34^\circ \). Then \( x_2=\frac{7500}{\tan(34^\circ)} \). Since \( \tan(34^\circ)\approx0.6745 \), we have \( x_2=\frac{7500}{0.6745}\approx11119.35 \) feet.

Step4: Calculate the distance \( d \)

The distance the plane travels \( d=x_1 - x_2 \). Substituting the values of \( x_1 \) and \( x_2 \): \( d = 21783.33- 11119.35=10663.98\approx10664 \) feet.

Answer:

10664