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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, 2019 at 12:07

3 Answers 3

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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, 2019 at 12:23
  • $\begingroup$ Thank you. I realised that only after typing my comment so I removed it. $\endgroup$
    – Vishnu
    Oct 22, 2019 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? $\endgroup$
    – Vishnu
    Oct 22, 2019 at 12:25
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    $\begingroup$ This is right my friend. $\endgroup$
    – Dr. Sonnhard Graubner
    Oct 22, 2019 at 12:26
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    $\begingroup$ I will think about your question. $\endgroup$
    – Dr. Sonnhard Graubner
    Oct 22, 2019 at 12:36
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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 $\theta_i$ ? $\endgroup$
    – Vishnu
    Oct 22, 2019 at 12:18
  • $\begingroup$ *calculate without using a calculator $\endgroup$
    – Vishnu
    Oct 22, 2019 at 12:54
  • $\begingroup$ I added a formula without trigonometry $\endgroup$
    – Vasily Mitch
    Oct 22, 2019 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. $\endgroup$
    – Vishnu
    Oct 22, 2019 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, 2019 at 16:53
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Use this theorem a Theorem for classifying triangles when given only the slopes of the equations

In order to know if a triangle is obtuse, acute or rectangle, one needs to know only the slopes of the equations of its sides.

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