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An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 750 m, horizontal distance 19.8 k...Asked by mohannad
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.)
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Answered by
KokkoDeRas
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
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
Answered by
Steve
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
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
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