calculating angle in circle












3














How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.



How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.



enter image description here






geometry trigonometry circle angle







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  • 1




    Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
    – Ramana Venkata
    Dec 27 '11 at 8:01










  • Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
    – J. M. is not a mathematician
    Dec 27 '11 at 9:29
















3














How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.



How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.



enter image description here






geometry trigonometry circle angle







share|cite|improve this question




















  • 1




    Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
    – Ramana Venkata
    Dec 27 '11 at 8:01










  • Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
    – J. M. is not a mathematician
    Dec 27 '11 at 9:29














3












3








3







How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.



How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.



enter image description here






geometry trigonometry circle angle







share|cite|improve this question















How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.



How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.



enter image description here






geometry trigonometry circle angle




geometry trigonometry circle angle






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edited Jul 20 '17 at 17:30









Martin Sleziak

44.7k7115270




44.7k7115270










asked Dec 27 '11 at 7:57









coure2011

200249




200249








  • 1




    Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
    – Ramana Venkata
    Dec 27 '11 at 8:01










  • Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
    – J. M. is not a mathematician
    Dec 27 '11 at 9:29














  • 1




    Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
    – Ramana Venkata
    Dec 27 '11 at 8:01










  • Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
    – J. M. is not a mathematician
    Dec 27 '11 at 9:29








1




1




Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01




Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01












Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29




Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29










6 Answers
6






active

oldest

votes


















7














I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:



angle = atan2(y2-y1, x2-x1)


(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.



The result comes out in radians, of course.






share|cite|improve this answer

















  • 1




    thanks for searching about me and giving me exactly what I was looking for.
    – coure2011
    Dec 28 '11 at 7:37



















6














HINT:



Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.






share|cite|improve this answer





























    5














    The angle can be unambiguously computed as
    $$
    theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
    $$
    This can be seen in the following diagram:



    arctan2






    share|cite|improve this answer





















    • This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
      – robjohn♦
      Dec 28 '11 at 23:27



















    1














    Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.



    Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.



    The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.



    Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$



    By solving for $cos A$ we get:
    $$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



    Which simplifies to:
    $$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



    Therefore the angle is:
    $$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$



    I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.



    Note: Please feel free to comment if I have made any careless mistakes.






    share|cite|improve this answer































      0














      You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
      At least end point of another line should also known to find the angle between those two lines in circle.






      share|cite|improve this answer

















      • 2




        It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
        – Henning Makholm
        Dec 28 '11 at 2:37



















      0














      Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.



      We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.






      share|cite|improve this answer























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        6 Answers
        6






        active

        oldest

        votes








        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7














        I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:



        angle = atan2(y2-y1, x2-x1)


        (which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.



        The result comes out in radians, of course.






        share|cite|improve this answer

















        • 1




          thanks for searching about me and giving me exactly what I was looking for.
          – coure2011
          Dec 28 '11 at 7:37
















        7














        I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:



        angle = atan2(y2-y1, x2-x1)


        (which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.



        The result comes out in radians, of course.






        share|cite|improve this answer

















        • 1




          thanks for searching about me and giving me exactly what I was looking for.
          – coure2011
          Dec 28 '11 at 7:37














        7












        7








        7






        I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:



        angle = atan2(y2-y1, x2-x1)


        (which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.



        The result comes out in radians, of course.






        share|cite|improve this answer












        I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:



        angle = atan2(y2-y1, x2-x1)


        (which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.



        The result comes out in radians, of course.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '11 at 2:34









        Henning Makholm

        238k16303538




        238k16303538








        • 1




          thanks for searching about me and giving me exactly what I was looking for.
          – coure2011
          Dec 28 '11 at 7:37














        • 1




          thanks for searching about me and giving me exactly what I was looking for.
          – coure2011
          Dec 28 '11 at 7:37








        1




        1




        thanks for searching about me and giving me exactly what I was looking for.
        – coure2011
        Dec 28 '11 at 7:37




        thanks for searching about me and giving me exactly what I was looking for.
        – coure2011
        Dec 28 '11 at 7:37











        6














        HINT:



        Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.






        share|cite|improve this answer


























          6














          HINT:



          Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.






          share|cite|improve this answer
























            6












            6








            6






            HINT:



            Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.






            share|cite|improve this answer












            HINT:



            Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '11 at 8:07









            davidlowryduda♦

            74.4k7117251




            74.4k7117251























                5














                The angle can be unambiguously computed as
                $$
                theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
                $$
                This can be seen in the following diagram:



                arctan2






                share|cite|improve this answer





















                • This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                  – robjohn♦
                  Dec 28 '11 at 23:27
















                5














                The angle can be unambiguously computed as
                $$
                theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
                $$
                This can be seen in the following diagram:



                arctan2






                share|cite|improve this answer





















                • This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                  – robjohn♦
                  Dec 28 '11 at 23:27














                5












                5








                5






                The angle can be unambiguously computed as
                $$
                theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
                $$
                This can be seen in the following diagram:



                arctan2






                share|cite|improve this answer












                The angle can be unambiguously computed as
                $$
                theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
                $$
                This can be seen in the following diagram:



                arctan2







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '11 at 23:20









                robjohn♦

                265k27303624




                265k27303624












                • This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                  – robjohn♦
                  Dec 28 '11 at 23:27


















                • This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                  – robjohn♦
                  Dec 28 '11 at 23:27
















                This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                – robjohn♦
                Dec 28 '11 at 23:27




                This is how to compute atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
                – robjohn♦
                Dec 28 '11 at 23:27











                1














                Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.



                Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.



                The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.



                Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$



                By solving for $cos A$ we get:
                $$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                Which simplifies to:
                $$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                Therefore the angle is:
                $$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$



                I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.



                Note: Please feel free to comment if I have made any careless mistakes.






                share|cite|improve this answer




























                  1














                  Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.



                  Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.



                  The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.



                  Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$



                  By solving for $cos A$ we get:
                  $$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                  Which simplifies to:
                  $$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                  Therefore the angle is:
                  $$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$



                  I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.



                  Note: Please feel free to comment if I have made any careless mistakes.






                  share|cite|improve this answer


























                    1












                    1








                    1






                    Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.



                    Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.



                    The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.



                    Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$



                    By solving for $cos A$ we get:
                    $$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                    Which simplifies to:
                    $$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                    Therefore the angle is:
                    $$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$



                    I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.



                    Note: Please feel free to comment if I have made any careless mistakes.






                    share|cite|improve this answer














                    Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.



                    Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.



                    The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.



                    Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$



                    By solving for $cos A$ we get:
                    $$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                    Which simplifies to:
                    $$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$



                    Therefore the angle is:
                    $$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$



                    I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.



                    Note: Please feel free to comment if I have made any careless mistakes.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 27 '11 at 11:47

























                    answered Dec 27 '11 at 11:19









                    E.O.

                    5,01452550




                    5,01452550























                        0














                        You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
                        At least end point of another line should also known to find the angle between those two lines in circle.






                        share|cite|improve this answer

















                        • 2




                          It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                          – Henning Makholm
                          Dec 28 '11 at 2:37
















                        0














                        You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
                        At least end point of another line should also known to find the angle between those two lines in circle.






                        share|cite|improve this answer

















                        • 2




                          It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                          – Henning Makholm
                          Dec 28 '11 at 2:37














                        0












                        0








                        0






                        You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
                        At least end point of another line should also known to find the angle between those two lines in circle.






                        share|cite|improve this answer












                        You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
                        At least end point of another line should also known to find the angle between those two lines in circle.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 27 '11 at 9:18









                        mevada.yogesh

                        1013




                        1013








                        • 2




                          It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                          – Henning Makholm
                          Dec 28 '11 at 2:37














                        • 2




                          It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                          – Henning Makholm
                          Dec 28 '11 at 2:37








                        2




                        2




                        It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                        – Henning Makholm
                        Dec 28 '11 at 2:37




                        It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
                        – Henning Makholm
                        Dec 28 '11 at 2:37











                        0














                        Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.



                        We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.






                        share|cite|improve this answer




























                          0














                          Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.



                          We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.






                          share|cite|improve this answer


























                            0












                            0








                            0






                            Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.



                            We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.






                            share|cite|improve this answer














                            Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.



                            We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 28 '11 at 22:02









                            Brandon Carter

                            7,21322538




                            7,21322538










                            answered Dec 27 '11 at 12:15









                            Tomarinator

                            1,32111023




                            1,32111023






























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