An airplane is flying horizontally on a flight path that has taken it directly over a radar tracking station.?

Question

The plane's altitude is 6 miles. Suppose that when the distance between the plane and the radar station is 10 miles, this distance is increasing at a rate of 500 miles per hour. Find the speed of the plane.

^this is all that is given and the following is my own attempt.
I'm in calculus and it's dealing with relative rate.

also ive gotten this so far but cant seem to get the right answer, although i dont know what it is. i am using an online homework machine so it tells me when its right or wrong.

given: when g=10miles dg/dt=500 mi/h
goal: find d(angle)/dt
Equation: Tan(angle) = 1/6 * g
soln: sec^2(angle) * d(angle)/dt = d/dt(1/6*g)
136/36 *d(angle)/dt = 1/6 * dg/dt
136/36 * d(angle)/dt = 500/6
d(angle)/dt = 500/6 * 36/136
d(angle)/dt = 3000/136

and that is not hte answer and i'm not understanding what i am doing wrong.

please help!!!

Answer 1

Unless you've been instructed otherwise, don't use an angle. Draw a picture of the situation, with the radar station on the ground and the plane in the air. Drop a vertical line segment from the plane to the ground. Connect this to the radar station with a horizontal line segment, and denote its length by x.

Then you have a right triangle, the hypotenuse of which is a line segment connecting the plane with the radar station (whose length you have denoted by g). The plane is always 6 miles above the ground, so the vertical line segment is always of length 6. Convince yourself that the rate at which x changes is the same as the plane's speed. Now, use the triangle to write a relationship between x and g so that you can find dx/dt knowing that dg/dt = 500 when g=10.

Answer 2

You are looking for the speed of the plane, not the rate of change of the angle with respect to time.

The speed should be 625 mph