Answer:
about 5.019 km
Step-by-step explanation:
You want the distance between two airplanes whose locations are ...
- 10.9 km W 33.9° S at 1123 m altitude
- 10.7 km W 21.9° S at 5601 m altitude
Coordinates
In (-x, -y, z) coordinates, the coordinates of the two airplanes in km are ...
Plane 1: (10.9·cos(33.9°), 10.9·sin(33.9°), 1.123) ≈ (9.047, 6.079, 1.123)
Plane 2: (10.7·cos(21.9°), 10.7·sin(21.9°), 5.601) ≈ (9.928, 3.991, 5.601)
The difference between these coordinates is ...
Plane 2 - Plane 1 = (9.928, 3.991, 5.601) -(9.047, 6.079, 1.123)
Plane 2 -Plane 1 = (0.881, -2.088, 4.478)
Distance
The distance between the planes is the root of the sum of the squares of these distances:
d = √(0.881² +(-2.088)² +4.478²) ≈ 5.019 . . . . km
The distance between the planes is about 5.019 km.
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Additional comment
We can use any convenient consistent coordinate system. Here, since both angles are measured CCW from west, we choose a coordinate system with the x-y plane rotated 180° from the usual position. The distances and relative angles are unchanged by this rotation.
Since we want the distance between planes at different altitudes, we have cast the problem as a 3D problem with axis measures in kilometers. Then the usual procedure for finding distances between points applies.
We have assumed both distance and angle measures are from the tower to the airplane. The wording of the problem is somewhat ambiguous. If one of these measures is from the airplane to the tower, then the distance between planes will be quite different.
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