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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.

diagram

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

Thanks

Colin

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    What do you mean by "bearing's"? Rays? Spheres?2011-05-09
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    On a sphere/globe or on a plane?2011-05-09
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    "bearing" - "direction"?2011-05-09
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    "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.2011-05-09
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    Points do not intersect. Rays from points or lines through points intersect.2011-05-10
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    Colin, but you could upload a pic to e.g. http://imgur.com and provide a link. Someone else can then include the pic into your post for you.2011-05-10
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    What are the bearings with respect to? I don't see a convenient reference that bearing 1 is 50 degrees from, nor that bearing 2 is 20 degrees from. Angles are between two things-I was taking up as the reference, but other conventions are possible.2011-05-10

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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 colin2011-05-10
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    @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.2011-05-10
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    @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.2011-05-10
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    @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...2011-05-10
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    @Ross: Given that bearings are angles clockwise from North, if the $\theta_i$ are bearings, then would we want $m_i=\tan(\frac{\pi}{2}-\theta_i)$?2011-05-10
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    @anyone: I presume the general consensus is that Ross's answer is correct? if so would it be possible for someone to rewrite it in a way a non maths person like myself could understand EG plain engish or a programming laungauge etc thanks colin2011-05-10
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    @Isaac: yes, that is right.2011-05-10
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    @meee I'm not sure how much easier it could be explained. You are given two equations with two unknowns (x3,y3) as an answer. This is trivial algebra; if you still don't know how to proceed, follow the simultaneous equations link and read the first three paragraphs on the wikipedia page. The only tricky part is the assumption that the bearings are measured in degrees from north -- you didn't specify how the bearings are measured, so you'll have to convert them if this assumption is incorrect.2011-05-10
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    Its the last bits I don't understand the $y 3 −y 1 =m 1 (x 3 −x 1 )$ and $y 3 −y 2 =m 2 (x 3 −x 2 )$2011-05-10
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    The point-slope equation for a line is $y−y_1 =m(x−x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. So I wrote the equations for each line and set $x$ and $y$ equal, as they must be at the point of intersection. The point of intersection is $(x_3,y_3)$2011-05-10
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    @meee have you followed the simultaneous equations link and read how to subtract one equation from the other to eliminate a variable? You need to eliminate y3 to solve for x3 and then eliminate x3 to solve for y3. See http://www.themathpage.com/alg/simultaneous-equations.htm#substitution for another step-by-step example and a graph.2011-05-10