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Given triangle ABC.


sinC = (sinA + sinB) / (cosA + cosB)


What details/characteristics can be determined about the shape of ABC (e.g. type of triangle) ?

 Mar 29, 2019


\(\textrm{Let}\quad\overline{BC}=a,\quad\overline{CA}=b,\quad\overline{AB}=c\\ \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R\quad (\textrm{R=radius of the circumscribed circle of triangle ABC})\\ \therefore sinA=\frac{a}{2R},\quad sinB=\frac{b}{2R},\quad sinC=\frac{c}{2R}\\ cosA=\frac{b^2+c^2-a^2}{2bc},\quad cosB=\frac{c^2+a^2-b^2}{2ca},\quad cosC=\frac{a^2+b^2-c^2}{2ab}\\ \therefore \frac{c}{2R}=\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}}\\ \quad c=\frac{2abc(a+b)}{a(b^2+c^2-a^2)+b(c^2+a^2-b^2)}=\frac{2abc(a+b)}{ab^2+c^2a-a^3+bc^2+a^2b-b^3}=\frac{2abc(a+b)}{c^2(a+b)-(a-b)^2(a+b)}=\frac{2abc}{c^2-(a-b)^2}\\ \therefore c^2-(a-b)^2=2ab\\ \quad c^2=2ab+(a-b)^2=a^2+b^2\\ \therefore \textrm{Triangle ABC is a right triangle that has angle C as the right angle.}\)

 Mar 30, 2019

Hi ThorganHarzard

Welcome to the   Web2.0calc   forum       laugh


That is a very impressive answer but you lost me at the ''=2R"   bit.    What has it to do with the circumscribed circle?


I am not questioning your knowledge, I just don't understand.

Melody  Mar 30, 2019
edited by Melody  Mar 30, 2019


The 2R comes from the standard proof of the sine rule.

Draw your triangle ABC with its circumscribing circle, radius R, centre O.

With the usual notation, angle BOC = 2A, so if you drop a perpendicular from O to BC you get a rt-angled triangle with

two sides R and a/2 and an angle A, from which a/2R = sin(A) and a/sin(A) = 2R.

Repeat for B and C and equate the three.


Here's an alternative answer to the original question.



 \(\displaystyle \sin(C)=2\sin(C/2)\cos(C/2)=2\sin(90-(A+B)/2)\cos(90-(A+B)/2)\\ =2\cos((A+B)/2)\sin((A+B)/2) \dots\dots\dots(1)\)



\(\displaystyle \frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}=\frac{2\sin((A+B)/2)\cos((A-B)/2)}{2\cos((A+B)/2)\cos((A-B)/2)}=\frac{\sin((A+B)/2)}{\cos((A+B)/2)}\dots\dots(2)\)


Equate (1) and (2), cancel and cross multiply,

\(\displaystyle \cos^{2}((A+B)/2)=1/2,\\\text{so }\cos((A+B)/2) = 1/\sqrt{2}\\\text{so }(A+B)/2=45\text{ deg}\\A+B=90\text{ deg}\\C=90\text{ deg}.\)



 Mar 31, 2019

Thanks Tiggsy.    cool


I shall learn from your answer properly in the morning :)

Melody  Mar 31, 2019

Welcome as a member Tiggsy.

I should have said this immediately but I did not realize straight away that you had joined!

I know you have been here forever but I finally talked you into becoming a member.  I am very pleased! 

Now I will be able to look back and find your fabulous answers (from this point forward)


Finally ... I have exhausted this question.  There was a lot for me to work through here. 

I do not remember seeing those formulas before that you used Tiggsy. Hopefully now they, and their proofs, are a little embedded into my brain.

Anyway, i finally fully understand both solutions so I am happy.    laugh cool laugh

Melody  Apr 11, 2019

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