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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 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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    @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