Determine the angles in the triangle for given equations of three sides of $\triangle$.

Problem is that if we go by finding the intersection of pair of lines, calculating the sides and finding the angles will be a very tedious process, so I was trying to find some alternative.

If we use formula $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$, then we are assuming that $\theta$ is acute, but it is not necessary as one angle in the triangle can be obtuse also.

So I tried something but I am not sure whether it will be true for all the triangles

enter image description here Let's try to find angle between AB and AC

For $\triangle AHI$,

$$\delta=\beta+\angle HAI$$ $$\angle CAB=\delta-\beta \text { where } \delta>\beta$$ $$\tan\angle CAB=\dfrac{m_{AC}-m_{AB}}{1+m_{AC}\cdot m_{AB}}$$

It can be observed that $m_{AC}>m_{AB}$

$$F'=180-\angle CBA+\beta$$ $$\angle CBA=180+\beta-F'$$ $$\tan \angle CBA=\dfrac{m_{AB}-m_{BC}}{1+m_{AB}\cdot m_{BC}}$$

It can be observed that $m_{AB}>m_{BC}$

$$F'=\delta+\angle ACB$$ $$\angle ACB=F'-\delta \text { where } F'>\delta$$ $$\tan\angle ACB=\dfrac{m_{CB}-m_{AC}}{1+m_{CB}\cdot m_{AC}}$$

It can be observed that $m_{AC}>m_{BC}$

So now after all this calculations, I tried to make some pattern out of it, but couldn't find anything significant. We cannot do these calculations all the time because these calculations are time taking and requires proper graph.

Any help or hints?

  • $\begingroup$ One problem with all of this is that there’s no value $m$ of the slope or a vertical line, so you would have to deal with that as a special case. Perhaps that doesn’t come up if the lines all have equations of the form $y=mx+b$. $\endgroup$
    – amd
    Jan 13, 2020 at 21:17

1 Answer 1


The angles between lines are those between the respective normal vectors, or their complements.
Therefore, given the equations of two sides, say $$\begin{aligned} ax+by+c&=0\\ a'x+b'y+c'&=0,\end{aligned}$$ we can find the angle $\varphi$ as $$\varphi=\arccos\left( \frac{aa'+bb'}{||(a,b)||.||(a',b')||}\right)\quad \text{or}\quad \varphi=\pi-\arccos\left( \frac{aa'+bb'}{||(a,b)||.||(a',b')||}\right)$$ (or a complement of those above to $2\pi,$ depending on chosen orientation of the normal vectors...)


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