# I have three random points, O, A, B, with these I can get angles &#x03B1;<!-- α --> and &#

I have three random points, O, A, B, with these I can get angles $\alpha$ and $\beta$
How can I get the angle C, or better, directional vector of c which will be evenly in the middle, between the other two angles. Keep in mind that the angles might be of any value.
This old answer almost gets me were I need to be, but how can I change the final formula presented in the answer $g=\text{arctan}\phantom{\rule{thinmathspace}{0ex}}\left(2\mathrm{tan}r\right)$ to work with non-right triangles? to work with non-right triangles?
I have found lots of results searching for this, but they don't seem to work in my case.
It looks like a good solution, but I don't know enough about trig to determine the part to change in the final formula to make it not right triangle dependant
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Jaruckigh
You can use simple vector addition. Say
$\stackrel{^}{a}=\frac{\stackrel{\to }{OA}}{|\stackrel{\to }{OA}|}\phantom{\rule{0ex}{0ex}}\stackrel{^}{b}=\frac{\stackrel{\to }{OB}}{|\stackrel{\to }{OB}|}$Since $|\stackrel{^}{a}|=|\stackrel{^}{b}|=1$, $\stackrel{^}{c}=\stackrel{^}{a}+\stackrel{^}{b}$ points along the bisector of $\mathrm{\angle }OAB$