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I have three random points in a 2D space, $A$, $B$ and $C$, how can I compute the angle between the abscisse and the bisector of $ABC$:

enter image description here

So I am looking for the angle $α$ (in red).

I'm using a computer and I can easily compute the azimuth $AZIMUTH_{BA}$ and $AZIMUTH_{BC}$

enter image description here

For this specific example I can do:

$$\frac{AZIMUTH_{BA}-AZIMUTH_{BA}}{2} + \frac{5\pi}{2} - AZIMUTH_{BA}$$

To get the angle $α$.

But how can I get a generalized formula to always get the right $α$ angle between $0$ and $2\pi$ ?

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  • $\begingroup$ Well, if you have the coordinates of of all the points, you could calculate the vector $$ \vec{v}_{\text{bisector}} = \frac{\vec{BA} + \vec{BC}}{2} $$ And then you could just use dot product to calculate the angle $\alpha$. $\endgroup$ – Matti P. Nov 13 at 10:40
  • $\begingroup$ The dot product between $v_{bisector}$ and the abscisse ? $\endgroup$ – obchardon Nov 13 at 10:47
  • $\begingroup$ Yeah, you could take $\vec{v}_{\text{bisector}} \cdot (- \hat{i})$. $\endgroup$ – Matti P. Nov 13 at 10:54
  • $\begingroup$ @MattiP. That only produces the bisector when $\lvert\overrightarrow{BA}\rvert=\lvert\overrightarrow{BC}\rvert$. If the vectors are of different lengths, you need to normalize them first. Try, for example, $A=(10,0)$, $B=(0,0)$ and $C=(0,1)$. $\endgroup$ – amd Nov 13 at 23:28

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