Blericker74

2022-07-09

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 $g=\text{arctan}\phantom{\rule{thinmathspace}{0ex}}(2\mathrm{tan}r)$ 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

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 $g=\text{arctan}\phantom{\rule{thinmathspace}{0ex}}(2\mathrm{tan}r)$ 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

Jaruckigh

Beginner2022-07-10Added 11 answers

You can use simple vector addition. Say

$\hat{a}=\frac{\overrightarrow{OA}}{|\overrightarrow{OA}|}\phantom{\rule{0ex}{0ex}}\hat{b}=\frac{\overrightarrow{OB}}{|\overrightarrow{OB}|}$Since $|\hat{a}|=|\hat{b}|=1$, $\hat{c}=\hat{a}+\hat{b}$ points along the bisector of $\mathrm{\angle}OAB$

$\hat{a}=\frac{\overrightarrow{OA}}{|\overrightarrow{OA}|}\phantom{\rule{0ex}{0ex}}\hat{b}=\frac{\overrightarrow{OB}}{|\overrightarrow{OB}|}$Since $|\hat{a}|=|\hat{b}|=1$, $\hat{c}=\hat{a}+\hat{b}$ points along the bisector of $\mathrm{\angle}OAB$

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