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.
Best Answer
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 $$\frac{3}{4}+ 1 -7 = \frac{3}{4} \times 1 \times -7$$