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assuming you're in a vacuum, and you have a beacon transmitting a signal, and you know its signal strength. How could you find out the bearing (direction) it's at, if you know its signal strength, and if which way you're facing is 0°?

I've tried figuring it out myself, but I just can't get the final stuff:

Assuming the signal strength is 100k

The inverse square law says

*intensity = 1/distance^2*Therefore,

*Distance = sqrt(1/Intensity)*

Therefore, distance (

*D*) would be:

*sqrt(1/100k) = 0.003162*

The distance is the Hypotenuse (

*a*)

Therefore,

*c*(the shortest side) =

*a/2*since in a right angled triangle, the hypotenuse is always double as long as the shortest side. Therefore,

*c = 0.001581*

b(The last side) is therefore

b

*sqrt(0.003162^2-0.001581^2) = 0.002738*

The area of the triangle is

*A = (b*c)/2 = 0.000002 (2 x 10^-6)*

The part I'm stuck on, is how to calculate angle

*ca*?

*sin (0.001581/0.003162)*Gets the angle

*0.479426°*Which doesn't seem right, since according to Google, it's supposed to be

*53.14°*

*(If this is in the wrong forum category, I'm sorry.)*