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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.