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Suppose we are given the equation of the sides of a triangle, how can we determine whether the triangle is obtuse angled or acute angled? In case of a right angled triangle, I would simply check whether the slopes $m_1$ and $m_2$ of any two lines follow the relation $m_1m_2=-1$. I know to find the angle between two intersecting lines with slopes $m_1$ and $m_2$ using the following formula:

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

The problem is, the above formula is helpful in finding only the positive values of the tangent function, or only for acute angles, due to the presence of the absolute value function.

Are there any other algorithm to distinguish acute angle triangles from obtuse angled triangles? Is it possible to use the same formula to find them?

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    $\begingroup$ Easiest way would be to look for the length of the sides and use the law of cosines $\endgroup$ – Peter Melech Oct 22 '19 at 12:07
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The easiest way to check is to see, whether the sum of angles calculated with your formula add up to $\pi$.

  1. Consider $y=1-2x$, $y=1+2x$, $y=0$. Angles $\theta_{1,2,3}=1.107, 1.107,0.9273$. The sum $\sum\theta_i=\pi$. The triangle is acute

  2. Consider $y=1-x/2$, $y=1+x/2$, $y=0$. Angles $\theta_{1,2,3}=0.4636, 0.4636, 0.9273$. The sum $\sum\theta_i=1.8545<\pi$. The triangle is obtuse.

Edit. If you want to cope without calculator, you can utilize the formula for tangent of sum of 3 angles and derive the following criteria. If $$ \left|\frac{1+m_1 m_2}{m_1-m_2}\right|\left|\frac{1+m_2 m_3}{m_2-m_3}\right| + \left|\frac{1+m_2 m_3}{m_2-m_3}\right|\left|\frac{1+m_3 m_1}{m_3-m_1}\right| + \left|\frac{1+m_3 m_1}{m_3-m_1}\right|\left|\frac{1+m_1 m_2}{m_1-m_2}\right| = 1, $$ then the triangle is acute. If it's not, it is obtuse.

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  • $\begingroup$ Thank you for your answer. But, how would I be able to calculate $\tan ^{-1}m$ for all values of $m$, in order to find $\theta_i$ ? $\endgroup$ – Guru Vishnu Oct 22 '19 at 12:18
  • $\begingroup$ *calculate without using a calculator $\endgroup$ – Guru Vishnu Oct 22 '19 at 12:54
  • $\begingroup$ I added a formula without trigonometry $\endgroup$ – Vasily Mitch Oct 22 '19 at 16:46
  • $\begingroup$ I think if $A+B+C=\pi$ then $\tan A +\tan B +\tan C= \tan A \tan B \tan C$, and so must be the new expression introduced. Or you may use the half angles which will work with the existing ones. $\endgroup$ – Guru Vishnu Oct 22 '19 at 16:51
  • $\begingroup$ My approach requires only $m_{1,2,3}$ and doesn't require finding the vertices first, lengths of sides second and cosine theorem next. $\endgroup$ – Vasily Mitch Oct 22 '19 at 16:53
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Hint: calculate $$a^2+b^2-c^2,a^2+c^2-b^2,b^2+c^2-a^2$$

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    $\begingroup$ It Comes from the Theorem of cosines: $$\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '19 at 12:23
  • $\begingroup$ Thank you. I realised that only after typing my comment so I removed it. $\endgroup$ – Guru Vishnu Oct 22 '19 at 12:23
  • $\begingroup$ If any one of the expressions turn out to be negative then it is a obtuse angled triangle, else it represents an acute angled triangle. If its zero then it is a right angled triangle. Could you please tell whether I am correct? $\endgroup$ – Guru Vishnu Oct 22 '19 at 12:25
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    $\begingroup$ This is right my friend. $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '19 at 12:26
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    $\begingroup$ I will think about your question. $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '19 at 12:36

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