Given a triangle and equation of all its sides, how do I understand if an internal angle is obtuse or acute? I am given the following question:-
In a triangle ABC, if the equation of sides AB, BC, and AC are $2x-y+4 = 0$, $x-2y-1=0$ and $x+3y-3 = 0$ respectively, then what is the tangent of the internal angle a?
So I used the following formula, for two lines with slopes $m_1$ and $m_2$, the acute angle between them is given by
$$\tan\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|$$
which evaluates to $\tan \theta = |7|$
But how do I know that the internal angle is obtuse or the acute one? This question has both +7 and -7 as choices (multiple correct question).
Drawing a rough diagram doesn't help either.
Is there any way to find out whether the angle is obtuse or acute beforehand....like an easier method than using the law of cosines by finding out the lengths of the sides, as that would make the question way too lengthy.
 A: Slope-wise
$AB:2x-y+4=0 \implies m_{AB}=2$, $BC:x-2y-1=0 \implies m_{BC}=1/2$, $m_{AC}=-1/3$
$$|tan B|=|(2-1/2)/(1+1)|=3/4, |\tan C|=|(1/2+1/3)/(1-1/6)|=1, |\tan A=|(2+1/3)/(1-2/3)|=7$$
In a Triangle ABC If $$|\tan A|+|tan B|+|\tan C|= |\tan A| \tan B| \tan C| ~~~(1)$$
then all angles are acute. Other wise the  |\tan*| will correspond to obtuse angle and it will be given $\pi-\tan^{-1}**.$
In this question, (1) is not satisfied as we have $$\frac{3}{4}+ 1 +7 \ne \frac{3}{4} \times 1 \times 7$$. So obtuse angle is $A=\pi-\tan^{-1}7.$
You will be pleased to see that
$$\frac{3}{4}+ 1 -7 = \frac{3}{4} \times 1 \times -7$$
A: The hint:
Solve three systems and you'll obtain: $$A\left(-\frac{9}{7},\frac{10}{7}\right)$$
$$B(-3,-2)$$ and $$C\left(\frac{9}{5},\frac{2}{5}\right).$$
Thus, $$AB=\frac{12\sqrt5}{7},$$ $$AC=\frac{36\sqrt{10}}{35}$$ and $$BC=\frac{12\sqrt5}{5}$$ and since $$AB^2+AC^2-BC^2=12^2\cdot5\left(\frac{1}{49}+\frac{18}{35^2}-\frac{1}{25}\right)<0,$$ we see that $$\measuredangle BAC>90^{\circ}.$$
Also, $$\cos\measuredangle BAC=\frac{12^2\cdot5\left(\frac{1}{49}+\frac{18}{35^2}-\frac{1}{25}\right)}{2\cdot\frac{12\sqrt5}{7}\cdot\frac{36\sqrt{10}}{35}}=-\frac{1}{5\sqrt2}$$ and since $$1+\tan^2\measuredangle BAC=\frac{1}{\cos^2\measuredangle BAC},$$
we obtain $$1+\tan^2\measuredangle BAC=50$$ or
$$\tan\measuredangle BAC=-7.$$
A: Let $t_i=|\tan\theta_i|$.
If $$t_1+t_2+t_3-t_1t_2t_3=0$$
then all three angles are acute. If not replace $t_1$ with $-t_1$. If the result is $0$ the angle $t_1$ is obtuse. Otherwise it is acute.
A: Although I did not double-check the math in Michael Rozenberg's answer, I completely agree with his approach.  I was therefore surprised that someone downvoted his answer and I upvoted it.  Just to clarify the problem:
All you have to do is solve for the actual lengths of each side of the triangle.  Then the preliminary question of whether a specific angle is obtuse is immediately answered by the Law of Cosines, since the cosine of an angle between 90 degrees and 180 degrees (exclusive) is negative and the cosine of an angle between 0 degrees and 90 degrees (exclusive) is positive.
A: 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.
