As usual, draw a diagram:
You have a triangle, ABC, with angle A = 30 deg.
The plane is at C
The station is at B
When the plane is overhead, BC = 1000, so AB = 1000sqrt(3)
Angle B subtends side b, and is sweeping through 90 deg at the moment in question
Let a be the distance from the station to the plane.
Now, at any time, b/sinB = a/sin30
b/2 = asinB
b'/2 = a' sinB + acosB B'
Now, when C is directly overhead, B = 90 deg.
b' is the speed of the plane, 640
320 = a' sin 90 + aB' cos 90
320 = a'
Makes sense, since at the moment in question, the hypotenuse of our triangle is twice the altitude of the plane. So, it will be increasing twice as fast as the distance (height) at that time.
shortly after taking off, a plane is climbing at an angle of 30° and traveling at a constant speed of 640 ft/sec as it passes over a ground radar tracking station. At that instant of time, the altitude of the plane is 1000 ft. How fast is the distance between the plane and the radar station increasing at that instant of time?
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