A flight traffic controller is watching two planes. One is 20.0 km away in a
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
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)
I mean west is - x direction :)
Thanks!
To find the distance between the two planes, let's start by calculating the x and y components of the positions of each plane.
For the first plane:
Ax = 20.0 km * sin(35.0º) = 20.0 km * 0.5736 = 11.472 km
Ay = 20.0 km * cos(35.0º) = 20.0 km * 0.8192 = 16.384 km
For the second plane:
Bx = 18.0 km * cos(75.0º) = 18.0 km * 0.2588 = 4.6584 km
By = 18.0 km * sin(75.0º) = 18.0 km * 0.9659 = 17.3878 km
Now we can find the total x and y distances between the two planes:
Rx = Ax + Bx = 11.472 km + 4.6584 km = 16.1304 km
Ry = Ay + By = 16.384 km + 17.3878 km = 33.7718 km
Finally, we can use the Pythagorean theorem to calculate the distance between the two planes:
R = √(Rx^2 + Ry^2) = √((16.1304 km)^2 + (33.7718 km)^2) ≈ 37.5366 km
Therefore, the two planes are approximately 37.5366 km apart.