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Simple problem.

  1. Sep 20, 2006 #1
    Okay, this should really be a fairly easy problem and my work is below this question:

    "A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is d1 = 384 m at 40° above the horizon. The airplane is tracked for another 123° in the vertical eastwest plane, the range at final contact being d2 = 770 m. Find the displacement of the airplane during the period of observation."


    Heres what I did:

    I first figured that the resultant R = d1-d2. So, I decided to break that into components.

    Rx = d1x - d2x ; Ry = d1y - d2y. Now, to find the magnitude, I would take the square root of the sum of Rx^2 and Ry^2. And the direction would be the inverse tangent of Ry/Rx.

    So, Rx = 384*cos(40) - (-770*Cos(57)). I took the other angle 180-123 = 57. So, Rx = 713.53m

    Ry = 384*sin(40) - 770*sin(57). Ry = -398.94 (Which I think is totally wrong).

    Now, the rest is pretty simple for me to do provided the above is correct, which I think is not.

    I solved this and got the answer for the magnitude as 817.48m. Unfortunately, the answer is wrong.

    Know what I am doing wrong here? Thanks :)

  2. jcsd
  3. Sep 20, 2006 #2


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    Homework Helper

    Take the triangle with sides d1, d2, and let's say d3, which is the displacement of the plane. The area of the triangle equals A = 0.5*d3*h (1), where h = d1*sin40. The area can be expressed as [tex]A=\frac{d_{1}^2 sin(123^{\circ}) sin \gamma}{2sin \alpha}[/tex], where [tex]\gamma[/tex] is the angle between d1 and d3, and [tex]\alpha[/tex] is the angle between d2 and d3. So, calculate the angles, get the area A, and you can get d3 from equation (1). I hope this works.
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