Problem

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The path of a satellite orbiting the earth causes it to pass directly over two tracking stations $A$ and $B$, which are $43 \mathrm{mi}$ apart. When the satellite is on one side of the two stations, the angles of elevation at $A$ and $B$ are measured to be $86.7^{\circ}$ and $83.2^{\circ}$, respectively.
NOTE: The picture is NOT drawn to scale.
How far is the satellite from station A?
distance from $A=$
$\mathrm{mi}$
How high is the satellite above the ground?
height =
$\mathrm{mi}$

Answer

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Answer

\(\boxed{\text{Therefore, the satellite is approximately 663.5 mi from station A and 662.7 mi above the ground.}}\)

Steps

Step 1 :Form a triangle with the satellite, station A, and station B. The angle at the satellite is the difference between the two given angles, which is \(86.7^\circ - 83.2^\circ = 3.5^\circ\).

Step 2 :Use the law of sines to find \(d_A\): \(\frac{d_A}{\sin(83.2^\circ)} = \frac{43 \mathrm{mi}}{\sin(3.5^\circ)}\).

Step 3 :Solve for \(d_A\) to get: \(d_A = \frac{43 \mathrm{mi} \cdot \sin(83.2^\circ)}{\sin(3.5^\circ)} \approx 663.5 \mathrm{mi}\).

Step 4 :Find the height of the satellite above the ground using the right triangle formed by the satellite, station A, and the point on the ground directly below the satellite. Use the sine of the angle of elevation at station A to find the height: \(\sin(86.7^\circ) = \frac{h}{d_A}\).

Step 5 :Solve for \(h\) to get: \(h = d_A \cdot \sin(86.7^\circ) \approx 663.5 \mathrm{mi} \cdot \sin(86.7^\circ) \approx 662.7 \mathrm{mi}\).

Step 6 :\(\boxed{\text{Therefore, the satellite is approximately 663.5 mi from station A and 662.7 mi above the ground.}}\)

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