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I have 2 points with bearing's coming from them, I need to calculate a 3rd point(intersection of bearing's from point 1,2) but am unsure of the maths required to do this. Could someone give me an example of how to do this.

Edit - The bearing's I am referring to are degree's from north. So I have point A + Point B and an imaginary line coming from from each point on the a particular bearing. I wish to know at what point the imaginary lines would cross.


To explain further -

I have point X,Y on a map and a bearing to an object(point 3) from point 1, I also have point 2 on the map and a bearing from point 2 to the same object (point 3) as point 1. what I need to do is to calculate the X,Y for point 3 using points 1,2. If it helps I would imagine the max distances between point 3 and points 1,2 would be a mile or so.

Maths was never my strongest point so if someone could explain how to do this in basic steps that would be great



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What do you mean by "bearing's"? Rays? Spheres? – El'endia Starman May 9 '11 at 22:05
On a sphere/globe or on a plane? – Isaac May 9 '11 at 23:00
"bearing" is usually, I think, direction given as an angle clockwise from North. Doing the problem, however, requires knowing whether we're working on a sphere/globe or on a plane. – Isaac May 9 '11 at 23:21
Points do not intersect. Rays from points or lines through points intersect. – Ross Millikan May 10 '11 at 0:32
Colin, but you could upload a pic to e.g. and provide a link. Someone else can then include the pic into your post for you. – t.b. May 10 '11 at 12:17

In the plane, if you are given two points $(x_1,y_1), (x_2,y_2)$ and the angles between the vertical and the vector to a third point $(x_3,y_3)$ as $\theta_1, \theta_2$ we have the slope of the line through $(x_1,y_1)$ and $(x_3,y_3)$ is $m_1=\tan(\theta_1+\frac{\pi}{2})$ and the slope of the line through $(x_2,y_2)$ and $(x_3,y_3)$ is $m_2=\tan(\theta_2+\frac{\pi}{2})$. Then $y_3-y_1=m_1(x_3-x_1)$ and $y_3-y_2=m_2(x_3-x_2)$. This gives two equations in two unknowns.

Added in response to comment: I used the point-slope form for the two lines. Some further discussion is at PurpleMath and at Mathwords. The slopes are given by your bearings. Normally the slope of a line is the tangent of the angle measured from the horizontal, but I assumed that your bearings are measured from the vertical (as they are usually taken from North). That accounts for the addition of $\frac{\pi}{2}=90^{\circ}$. Given two lines, the intersection is found by finding a point $(x_3,y_3)$ that lies on both. This gives two simultaneous equations to solve for the two coordinates.

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Could you explain in terms an idiot could understand thanks colin – meee May 10 '11 at 9:18
@meee Colin, may I suggest making a drawing of what you want to do; that may vastly help us to explain the mathematics you need. – J. M. May 10 '11 at 9:49
@J.M. I'm pretty sure Ross Millikan interpreted it right. You have two points and two slopes given as angles from the vertical ($\pi/2$) and want the intersection between the two lines thus determined. – Ben Alpert May 10 '11 at 14:52
@Ben: I know Ross did it correctly; my point was that Ross's solution might have been more transparent to Colin had Colin sketched out his situation first... – J. M. May 10 '11 at 15:09
@Ross: Given that bearings are angles clockwise from North, if the $\theta_i$ are bearings, then would we want $m_i=\tan(\frac{\pi}{2}-\theta_i)$? – Isaac May 10 '11 at 16:35

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