# Simple problem.

1. Sep 20, 2006

### Double D Edd

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 :)

Edd.

2. Sep 20, 2006

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 $$A=\frac{d_{1}^2 sin(123^{\circ}) sin \gamma}{2sin \alpha}$$, where $$\gamma$$ is the angle between d1 and d3, and $$\alpha$$ 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.