The radar operator at an airport control tower locates two planes flying toward the airport at the same altitude. one plane is 120km away from the airport at a bearing of north 70 degree east. the other is 180km away, on a bearing of south 55 degree east. how far apart are the planes?

In order to answer this question, we find first the x and y components of the distances given.
 1st plane:
                      x-component: x = (120 km)(sin 70°) = 112.76 km
                      y-component: y = (120 km)(cos 70°) = 41.04 km
2nd plane:
                      x-component: x = (180 km)(sin 55°) = 147.45 km
                      y-component: y = (180 km)(cos 55°) = -103.24 km (negative because it is in the south)

The distance between these planes is calculated through the equation
                                   d = √(x₂ - x₁)² + (y₂ - y₁)²
Substituting the calculated values
                               d = √(112.76 - 147.45)² + (41.04 - (-103.24))²
                                   d = 148.39 km
Thus, the distance between these planes is approximately 148.39 km