How to determine whether a triangle is obtuse angled or not from the equation of its sides?

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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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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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Hint: calculate $$a^2+b^2-c^2,a^2+c^2-b^2,b^2+c^2-a^2$$

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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.