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This is the formula for finding the angles between two straight lines:

$$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

$$\implies \tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$$

$$\implies \theta=\pm\arctan\left(\frac{m_1-m_2}{1+m_1m_2}\right)$$

In LHS, only positive values of $\theta$ can be inputted.

Now, according to my book and this derivation, the $\pm$ has been included to include both the acute and obtuse angles between the straight lines. However, according to @AmanKushwaha, the $\pm$ sign has been included to include the positive (anticlockwise) and negative (clockwise) acute angles between the two straight lines. Who is correct?

Moreover, can't the acute angle measured in the clockwise direction also represent the obtuse angle formed between any two straight lines?

Thomas Andrews
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tryingtobeastoic
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  • the acute/obtuse interpretation is the correct one – hellofriends Sep 08 '21 at 05:08
  • First thing, $|x|$ is not same as $\pm x$.Perhaps, you meant this formula $ \small \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \tag{1} \label{1}$ Now what I actually said- You get two values of $m_2$ for a given $m_1$ and an acute inclination angle $\theta$ from $(\ref{1})$ because $\theta$ can be measured in two different directions namely, clockwise and anticlockwise, with respect to the given line with slope $m_1$, resulting in two different equations of lines both making same inclination angle $\theta$. – Aman Kushwaha Sep 08 '21 at 05:28
  • @Amankushwaha I edited the formula; is it okay now? – tryingtobeastoic Sep 08 '21 at 06:02
  • "In LHS, only positive values of θ can be inputted. " Actually, $\tan{\theta}$ is positive and $\theta$ is acute. So the correction would be "In LHS, θ can take only acute angles". – Aman Kushwaha Sep 08 '21 at 06:28
  • I'd suggest you to edit back the question to what you've asked originally otherwise the question asking for $\theta$ being acute or obtuse won't make any sense since now you should've understood that $\theta$ is always the acute angle between the lines in $\small \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$ and for given $m_1$ and $m_2$, you get exactly one $\theta$ and that will always belong to $\left[0,\frac{\pi}{2}\right)$ but for given acute $\theta$ and $m_1$ , you'll get two values of $m_2$. This equation $\small \tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$, however, yields.... – Aman Kushwaha Sep 08 '21 at 07:18
  • ... two different values of $\theta$ as mentioned in the answer below by @GregMartin . Read this comment I added in the answer of your previous question: https://math.stackexchange.com/questions/4244257/intuition-behind-getting-two-straight-lines-as-result?noredirect=1#comment8822539_4244391 – Aman Kushwaha Sep 08 '21 at 07:21
  • Both interpretations are correct, and they are actually identical. – Trebor Sep 08 '21 at 14:28

1 Answers1

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Note that the equation can be rewritten as $$ \theta = \arctan\biggl( \pm\frac{m_1-m_2}{1+m_1m_2} \biggr). $$ The answer to your question depends on the convention being used for the values of the arctangent function:

  • If one declares that $\arctan x$ always lies between $-\frac\pi2$ and $\frac\pi2$, then the formula yields the positive and negative acute angles formed by the two lines;
  • If one declares that $\arctan x$ always lies between $0$ and $\pi$, then the formula yields the acute and obtuse angles made by the two lines.
Greg Martin
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  • Thanks for your great answer! Could you please look into the linked question in my question? It'd be of great help: https://math.stackexchange.com/a/4244391/768162 – tryingtobeastoic Sep 08 '21 at 05:20