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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.

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  • $\begingroup$ It's not so hard. It takes 10 minutes of work. See my solution. I wait another solutions, which will take less minutes. By the way, for understand that $\measuredangle BAC>90^{\circ}$ we can take less of time. $\endgroup$ – Michael Rozenberg Mar 27 '20 at 15:26
  • $\begingroup$ @MichaelRozenberg This is a question from an examination in India(for std. 12 students), we have on average 2-3 mins per question....no calculators allowed $\endgroup$ – Techie5879 Mar 27 '20 at 15:30
  • $\begingroup$ Also, we can draw this triangle and we can see that the angle is obtuse. But it's not so math of course. $\endgroup$ – Michael Rozenberg Mar 27 '20 at 15:32
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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 $$\frac{3}{4}+ 1 -7 = \frac{3}{4} \times 1 \times -7$$

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  • $\begingroup$ Why just $\angle BAC$ is obtuse? Why not $\angle ABC$, for example? $\endgroup$ – Michael Rozenberg Mar 27 '20 at 15:35
  • $\begingroup$ Because if you makr $3/4$ negative the identity will not be satisfied as happened in the last eq. by making 7 negative. So A is obtuse A=\pi-\tan^{-1}7$ $\endgroup$ – Z Ahmed Mar 27 '20 at 15:41
  • $\begingroup$ I agree. It's really nice. Thank you! $\endgroup$ – Michael Rozenberg Mar 27 '20 at 15:44
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    $\begingroup$ It is a pleasure to hear this from you. It is a solid method. $\endgroup$ – Z Ahmed Mar 27 '20 at 15:59
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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 $$\tan\measuredangle BAC=-7.$$

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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.

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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.

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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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