# Telling the greatest angle of the triangle when slopes are given

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.

• 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 $\tan^{-1}$) – Jean Marie May 21 '19 at 7:58

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 as the answer to this question.