Question

the path of a satellite orbiting the earth causes it to pass directly over two tracking stations a and b, which are 69 km apart. when the satellite is on one side of the two stations, the angles of elevation at a and b are measured to be 86.9° and 84.9°, respectively.
note: the picture is not drawn to scale.
how far is the satellite from station a?
distance from a = km
how high is the satellite above the ground?
height = km
enter your answer as a number; your answer should be accurate to 2 decimal places.
question help: video

Explanation:

Step1: Find the angle at the satellite

The sum of angles in a triangle is \(180^{\circ}\). Let the angle at the satellite be \(C\). Then \(C = 180-(86.9 + 84.9)=8.2^{\circ}\)

Step2: Use the Law of Sines to find the distance from station \(A\) (let it be \(a\))

By the Law of Sines \(\frac{a}{\sin B}=\frac{c}{\sin C}\), where \(c = 69\) km (distance between \(A\) and \(B\)), \(B = 84.9^{\circ}\), \(C=8.2^{\circ}\)
\(a=\frac{69\times\sin84.9^{\circ}}{\sin8.2^{\circ}}\)
\(\sin84.9^{\circ}\approx0.997\), \(\sin8.2^{\circ}\approx0.142\)
\(a=\frac{69\times0.997}{0.142}\approx483.66\) km

Step3: Find the height \(h\) of the satellite

We know that \(\sin86.9^{\circ}=\frac{h}{a}\) (using the right - triangle formed with station \(A\), the point on the ground directly below the satellite and the satellite)
Since \(a\approx483.66\) km, \(h = a\times\sin86.9^{\circ}\)
\(\sin86.9^{\circ}\approx0.998\)
\(h = 483.66\times0.998\approx482.79\) km

Answer:

distance from \(A\approx483.66\) km, height\(\approx482.79\) km