# Trigonometry - Bearings

I have a Trigonometry problem I just can't seem to grasp: Given a coordinate, find a bearing to reach that coordinate.

What I have tried thus far:

I have another problem in the same section that gives the coordinate to be reached as (5,0). In order to solve this, I just thought of how it would be graphed on the Cartesian Coordinate Plain: Five positive units right, no units up. This would be on the x-axis and, being that coordinates start at the y-axis, it would be N 90 E (90 degrees to the right of the y-axis)

However, when I get something such as (3,3) or (2,-2), I have no idea where to start.

Any help is greatly appreciated.

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This may help you: movable-type.co.uk/scripts/latlong.html – NoChance Feb 4 '12 at 18:52
I have found this, but I need a way to do it like I said above. That could include just using the Trigonometric functions or logic. – nmagerko Feb 4 '12 at 19:03
Is this problem on the plane (which the question seems to indicate), or on the sphere (which the latlong.html link assumes)? – robjohn Feb 4 '12 at 20:26
Sorry I did not know about degree of measurement. You should follow robjohn answer as it looks accurate. – NoChance Feb 4 '12 at 20:42
Sounds good, Emmad. This is on a plane, Robjohn. You are correct. – nmagerko Feb 4 '12 at 20:51

Assuming that your bearing is measured east of north (right of the $y$-axis), you can use $$\text{bearing}=2\;\tan^{-1}\left(\frac{x}{y+\sqrt{x^2+y^2}}\right)\tag{1}$$ $\text{bearing}<0^\circ$ means west of north.
$\text{bearing}>90^\circ$ means south of east.
$\text{bearing}<-90^\circ$ means south of west.
Formula $(1)$ only fails when $y\le0$ and $x=0$ (division by $0$); in that case, $\text{bearing}=180^\circ$, due south.