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