An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 850 m, horizontal distance 20.0 km, and 24.5° south of west. The second aircraft is at altitude 1100 m, horizontal distance 18.0 km, and 16.0° 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.)

i do not know if this is process you'll be fine. after finding the spatial coordinates of tha sobstitute in the distans equation between two points:

first aircraft angle a1=24.5*2Pg/360=0.43Rad
second aircraft angle
a2=16.0*2Pg/360=0.28Rad

first aircraft d1=20000m
x1=d1*sen(a1) y1=d1*cos(a1) z1=850m

second aircraft d2=18000m
x2=d2*sen(a2) y2=d2*cos(a2) z2=1100m

distance(1,2)=sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)=3485,3m
........perhaps

using x,y,z as coordinates, in km,

p1 = (20cos24.5,20sin24.5,0.85)
= (18.199,8.294,0.85)

p2 = (18cos16,18sin16,1.1)
= (-17.303,-4.961,1.1)

d^2 = (18.199-17.303)^2 + (8.294-4.961)^2 + (.85-1.1)^2
d = 3.460km

To find the distance between the two aircraft, we can use the concept of vector addition. First, we need to break down the altitude and the horizontal distance of each aircraft into their respective components.

For the first aircraft:
- Altitude: 850 m (let's call this vector A)
- Horizontal distance: 20.0 km at 24.5° south of west (let's call this vector B1)

For the second aircraft:
- Altitude: 1100m (let's call this vector C)
- Horizontal distance: 18.0 km at 16.0° south of west (let's call this vector B2)

To find the components of vectors B1 and B2, we need to use trigonometry.

For vector B1:
- The horizontal component (B1x) can be found using the cosine of the given angle: B1x = 20.0 km * cos(24.5°)
- The vertical component (B1y) can be found using the sine of the given angle: B1y = 20.0 km * sin(24.5°)

For vector B2:
- The horizontal component (B2x) can be found using the cosine of the given angle: B2x = 18.0 km * cos(16.0°)
- The vertical component (B2y) can be found using the sine of the given angle: B2y = 18.0 km * sin(16.0°)

Now, we can calculate the final position vector for each aircraft by adding the respective altitude vector to the horizontal distance vector components.

For the first aircraft:
- Position vector A = (0, -850m, 0)

For the second aircraft:
- Position vector C = (0, -1100m, 0)

- Position vector B1 = (B1x, B1y, 0)
- Position vector B2 = (B2x, B2y, 0)

Finally, to find the distance between the two aircraft, we can subtract the position vectors and calculate the magnitude of the resulting vector.

Distance between the two aircraft = |(C - A) - (B2 - B1)|

where "-" represents vector subtraction and "|" denotes the magnitude of the resulting vector.

Calculate (C - A):
(C - A) = (0, -1100m, 0) - (0, -850m, 0)

Calculate (B2 - B1):
(B2 - B1) = (B2x, B2y, 0) - (B1x, B1y, 0)

Finally, calculate the magnitude of the final vector to find the distance between the two aircraft.