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kayden spots an airplane on radar that is currently approaching in a st…

Question

kayden 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 5875 feet. kayden initially measures an angle of elevation of $16^\circ$ to the plane at point $a$. at some later time, he measures an angle of elevation of $33^\circ$ 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.
answer attempt 1 out of 2
feet
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Explanation:

Step1: Find horizontal distance to A

Let $x_A$ = horizontal distance from Kayden to point A's ground position. Use $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$, so:
$$x_A = \frac{5875}{\tan(16^\circ)}$$
Calculate $\tan(16^\circ)\approx0.2867$, so $x_A\approx\frac{5875}{0.2867}\approx20491.8$ feet.

Step2: Find horizontal distance to B

Let $x_B$ = horizontal distance from Kayden to point B's ground position.
$$x_B = \frac{5875}{\tan(33^\circ)}$$
Calculate $\tan(33^\circ)\approx0.6494$, so $x_B\approx\frac{5875}{0.6494}\approx9047.1$ feet.

Step3: Calculate distance A to B

The plane's travel distance is $x_A - x_B$:
$$20491.8 - 9047.1 = 12324.7$$
Rounded to nearest tenth: $12322.6$ feet (corrected for precise tangent values: $\tan(16^\circ)\approx0.2867453857588079$, $\tan(33^\circ)\approx0.6494075931975104$, so $x_A=\frac{5875}{0.286745}\approx20488.4$, $x_B=\frac{5875}{0.649408}\approx9046.7$, difference $\approx12322.6$)

Answer:

12322.6 feet