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When coordinates or position vectors (2D or 3D) of vertices A,B,C of a triangle ABC or its side vectors AB, BC, CA are given we can find the largest angle of $\Delta$ ABC by finding the angle against the largest side employing the cosine law. But when slopes $m_{AB}, m_{BC}, m_{CA}$ of the sides of $\Delta$ ABC arte given it becomes confusing to tell the largest angle confidently.

Can some one help me with a systematic approach in such situations? I have 5 questions in this regard.

Tell name and the value of the greatest angle of $\Delta$ ABC, if the slopes of AB, BC, CA are (i) $3/2, 1/3, 1/2$, (ii) $-3, -1/3, 3/2,$, (iii) $1/3, 7/3, 8$, (iv) $5/4, 7/3, 1/2$; and (v) Given the lines AB: $x+2y=3$, BC: $2x+y+4=0$ and CA: $y=mx-2$, find all values of $m$ such that the $\angle C$ of the triangle is obtuse.

Thanks in advance.

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  • $\begingroup$ Convert slopes into $(-\pi/2,\pi/2)$ polar angles (relative to an horizontal reference) by taking their atan$(m_{AB})$ etc... (atan is the correct naming for $\tan^{-1}$) $\endgroup$ – Jean Marie May 21 '19 at 7:58
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You are right, the issue of finding the greatest angle of the $\Delta ABC$ becomes interesting and confusing. Here is an approach to ward off any ambiguity or confusion. when slopes $m_{AB}, m_{BC}$, and $m_{CA}$ of three lines AB, BC, CA are given. One should find $$\tan B= \frac {m_{AB}-m_{BC}}{1+m_{AB}~m_{BC}} ~~~(1)$$ etc. cyclically as $\angle B (AB,BC), ~\angle C (BC,CA)$, and $\angle A (CA,AB)$. Two cases arise here.

Case-1: When all three of tans are positive or all three of them are negative, the triangle is acute angle triangle satisfying the marvelous condition that $$\tan A+ \tan B+ \tan C= \tan A~\tan B~\tan C,~~~(2)$$ The greatest of tans (say $\tan C$) will yield the greatest acute angle of $\Delta ABC$ as $C=\tan^{-1} |\tan C|$.

Case-2: Two of the tans are positive and one of them is negative (say, $\tan C$). Or two of them are negative and one of them is positive (say, $\tan C$). Then $\angle C=\pi- \tan^{-1}|\tan C|$ is the greatest angle of the triangle which is obtuse. However, in either of the sub-cases, Eq.(2) will be satisfied.

In your question (i) $\tan B=7/9, \tan C=-1/2, \tan A=-4/7$ the case-2 applies and hence the largest angle is an obtuse angle $B= \pi-\tan^{-1} 7/9$. The the question (ii) corresponds to case-1, as $\tan B =-4/3, \tan C=-11/3, \tan A=-9/7$ (all negative) so the greatest angle is $C= \tan^{11/4}$ and ABC is an acute angle triangle.

The question (v) is more interesting. Here, $\tan B=3/4$ which is positive, we seek here the case II discuuesed above according to this $\tan C<0$ and $\tan A>0$ (two of tans are positive and one is negative). Consequently, we get $$\tan C= \frac{-2-m}{1-2m} <0 ~ \mbox{and} ~\tan A= \frac{m+1/2}{1-m/2} >0. ~~~(3)$$ Finally, the overlap of inequations in (3) gives $-1/2 <m <1/2$ as the answer to this question.

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