An air-traffic controller notices two aircraft on his radar screen. The first is at altitude 850 m, horizontal distance 19.6 km, and 27.0° south of west. The second aircraft is at altitude 1100 m, horizontal distance 17.4 km, and 18.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.)

To find the distance between the two aircraft, we can use the concept of vector addition.

First, let's convert the given information into vectors. Since the x-axis is west and the y-axis is south, we need to use the negative sign for the magnitude of the vector in the respective direction.

For the first aircraft:
Magnitude (magnitude of the horizontal distance): 19.6 km
Direction (south of west): 27.0°
Let's calculate the x-component and y-component of this vector:
x-component = 19.6 km * cos(27.0°) = 16.992 km (west)
y-component = -(19.6 km * sin(27.0°)) = -9.255 km (south)

For the second aircraft:
Magnitude (magnitude of the horizontal distance): 17.4 km
Direction (south of west): 18.0°
Let's calculate the x-component and y-component of this vector:
x-component = 17.4 km * cos(18.0°) = 16.553 km (west)
y-component = -(17.4 km * sin(18.0°)) = -5.049 km (south)

Now we have two vectors: A = (16.992 km, -9.255 km) and B = (16.553 km, -5.049 km).

To find the distance between the two aircraft, we can use the distance formula:
distance = √((x2 - x1)^2 + (y2 - y1)^2)

Let's plug in the values:
distance = √((16.553 km - 16.992 km)^2 + (-5.049 km - (-9.255 km))^2)
distance = √((-0.439 km)^2 + (4.206 km)^2)
distance = √(0.193 km^2 + 17.677 km^2)
distance = √17.870 km^2
distance = 4.226 km

Therefore, the distance between the two aircraft is approximately 4.226 km.