A flight traffic controller is watching two planes. One is 20.0 km away in a

Question

direction 35.0º west of north at an altitude of 1650 m. The other is 18.0 km
away in a direction 75.0º north of east at an altitude of 1420 m. How far apart are the two planes?

Ax = 20.0km*sin(35.0º)
Ay = 1.650km or 20.0km*cos(35.0º)?

Bx = 18.0km*cos(75.0º)
By = 1.420km or 18.0km*sin(75.0º)?

Rx = Ax + Bx
Ry = Ay + By

R = √(Rx^2 + Ry^2)= distance between the two airplanes

Answer 1

West is - y direction

Ax = -20.0km*sin(35.0º) NOTE NEGATIVE
Ay = 1.650km or 20.0km*cos(35.0º)?

Bx = 18.0km*cos(75.0º)
By = 1.420km or 18.0km*sin(75.0º)?

Rx = Ax + Bx
Ry = Ay + By

and Rz = 1.650 - 1.420

R = sqrt (Rx^2 + Ry^2 + Rz^2)

Answer 2

I mean west is - x direction :)

Answer 3

Thanks!