Question

The figure below shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A and B, which are 68 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 83.9 degrees and 86.2 degrees respectively. How far is the satellite from station

A and how high is the satellite above the ground? Round answers to the nearest whole mile.

Answer 1

Explanation:

let the third vertex=C

180-86.2=93.8°

angle near the satellite=180-(83.9+93.8)=180-177.7°=2.3°

angles of triangle are 83.9°,93.8°,2.3°

by sine formula

`(AC)/(Sin~93.8) =(AB)/(Sin 2.3) \\(AC)/(Sin(180-93.8)) =(68)/(Sin~2.3) \\(AC)/(Sin~86.2) =(68)/(Sin~2.3) \\AC=(Sin~86.2)/(Sin~2.3) *~68\\AC \approx 1690.69\\AC\approx1691~mile`

Answer 2

The satellite is 640 miles from station A and 1020 miles above the ground.

In the triangle formed by the satellite, station A, and point B on the ground directly below the satellite, we have the following information:

Angle A = 83.9 degrees

Angle B = 86.2 degrees

Side AB = 68 miles

We want to find Side AC (the distance from the satellite to station A) and Side BC (the height of the satellite above the ground)

Using the law of cosines, we can find Side AC:

AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(A)

Substituting in the known values, we get:

AC^2 = 68^2 + BC^2 - 2*68*BC*cos(83.9)

Simplifying and solving for BC, we get:

BC = (68^2 + AC^2) / (2*68*cos(83.9))

Using the law of sines, we can find Side BC:

BC/sin(A) = AB/sin(B)

Substituting in the known values, we get:

BC/sin(83.9) = 68/sin(86.2)

Simplifying and solving for BC, we get:

BC = 68*sin(83.9) / sin(86.2)

Equating the two expressions for BC, we get:

(68^2 + AC^2) / (2*68*cos(83.9)) = 68*sin(83.9) / sin(86.2)

Solving for AC, we get:

AC = 640 miles

Now that we know AC, we can plug it back into the expression for BC to find BC:

BC = (68^2 + 640^2) / (2*68*cos(83.9))

BC = 1020 miles

Therefore, the satellite is 640 miles from station A and 1020 miles above the ground.