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This question is linked to Can known object be used to back-calculate my location? (been almost a month, figured it would be best to start a new question.)

I have a map, and I know which way true north is. I also have marked a known location on that map. I know my distance from that location, as well as my bearing relative to the location. Based on that info, I've been trying to come up with a formula for how to back-calculate MY location. I've been messing around with some trig formulas to get it, but I just don't understand how I would get my location. Like I said in the linked question, my distance from that known location obviously means how far I am. But I'm confused on the angles part, and any help would be great! A drawing might best help. Thank you.

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Let's start with this image I found that is kind of the reverse of your situation. In this image the coordinates are known and the bearing is calculated. enter image description here

Let's say $(x_0,y_0)$ is the location you know. You can set up a right triangle using your known bearings to figure out what amount you need to add or subtract to your $x$ value and your $y$ value.

EDIT: So, you know $(x_0,y_0)$ and your bearing from $(x_0,y_0)$. Assuming you are expressing bearing as the angle from NORTH going clockwise, you can express your coordinates as $$(x_0+d(\sin(\beta)),y_0+d(\cos(\beta)))$$

Note that $\sin(\beta)$ goes with $x$ instead of the standard $y$ you might be used to with the unit circle. This is because normal degrees start with EAST on the unit circle, but these bearings start with zero at NORTH.

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  • $\begingroup$ Thanks for the response. Wouldn't I have to know the x,y values of my location first though to do that? Or, would it be something like doing the distance formula backwards? $\endgroup$ – pfinferno Dec 18 '14 at 17:48
  • $\begingroup$ I'll edit my answer... $\endgroup$ – turkeyhundt Dec 18 '14 at 17:52
  • $\begingroup$ Okay so, bearing is sort of set up like a speedometer on a car? 0 on the left, 360 on the right? Maybe if we used some numbers I could see it a little better. Let's say my x,y coordinates of the known location are (2, 4) and the bearing is 30 degrees. The hypotenuse is 20. $\endgroup$ – pfinferno Dec 18 '14 at 18:05
  • $\begingroup$ Bearing is usually 0 = NORTH, 90 = EAST, 180 = SOUTH, 270 = WEST, 360 = NORTH again. In your example, you are 20 from (2,4) at a bearing of 30 degrees, I'd say your location is $(2+20(\sin(30^{\circ})),4+20(\cos(30^{\circ})))$ = $(12,21.32)$ $\endgroup$ – turkeyhundt Dec 18 '14 at 18:16
  • $\begingroup$ Oh! Alright this is starting to make sense. Just one question about the bearing here. From what I learned, bearing is the direction you are facing. So if I say my bearing relative to the object is 30 degrees, is that 30 degrees I'm facing sort of north-east away from the object? As in, I need to go northeast to reach that object? $\endgroup$ – pfinferno Dec 18 '14 at 18:20

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