help

Am confused by question

What do you mean by sides OPPOSITE of A, B, C? is there a picture?

In triangle ABC we have (a+b+c)(a+b-c)=3ab, where a, b and c are

the sides opposite A, B and  C respectively.

What is the degree measure of \(\angle\ C=\angle\ \gamma\)  ?

\(\color{BrickRed}(a+b+c)(a+b-c)=3ab\\ a²+ab-ac+ab+b²-bc+ac+bc-c²-3ab=0\\ a²-ab+b²=c² \)

\(c²=a²+b²-\color{blue}ab\\ c²=a²+b²-\color{blue}2ab \cdot cos (\gamma)\ \color{red}(was\ wrong\ before:\ -2bc\cdot cos(\gamma))\\ 2ab\cdot cos(\gamma)=ab\\ cos(\gamma)= \frac{ab}{2ab}\\ cos(\gamma)=\frac{1}{2}\)      

was corrected by me

\(\gamma=arc\ cos\ ( \frac{1}{2})\)

\(C=\gamma=60°\)

be inapplicable:

[If it is possible to determine the ratio ( \(\large \frac{a}{c}\) \( |\ c\ge 2a\) ) from equation\( (a+b+c) (a+b-c) = 3ab\), \(\angle\ \gamma\) can also be determined with the equation \(\gamma=arc\ cos\ ( \frac{a}{2c})\).]

Danke CPhill!

  !

( a + b + c) (a + b - c)  = 3ab

[ (a + b) + c ]  [(a + b) - c ]  = 3ab

(a + b)^2  - c^2  = 3ab

a^2 + 2ab + b^2 - c^2 = 3ab

a^2 + b^2   =  ab + c^2     (1)

Using the Law of Cosines

c^2  = [a^2 + b^2] - 2(a)(b) cos C      (2)

Sub (1)  into (2)    and we have that

c^2  = [ ab + c^2]  - (2ab) cos C

c^2  = ab + c^2 - (2ab) cos C       subtract c^2 from both sides

0   =  ab - 2ab cos C        subtract ab from both sides

-ab  = - 2ab cos C            divide both sides by -2ab

1/2  =  cos C

arccos (1/2)  = C  =  60°

Thanks to asinus for the hint of applying the Law of Cosines  !!!