The Three Sides

We should know the cosine rule which states if the triangle ABC has sides $BC=a,AC=b,AB=c$ , then $\cos C =\frac{a^2+b^2-c^2}{2ab}$ . Given that ---

$(a+b+c)(a+b-c)=3ab$

$\implies (a+b)^2-c^2=3ab$ [since $a^2-b^2=(a+b)(a-b)$ ]

$\implies a^2+2ab+b^2-c^2=3ab$

$\implies a^2+b^2-c^2=ab$

$\implies \frac{a^2+b^2-c^2}{ab}=1 \implies \frac{a^2+b^2-c^2}{2ab}=\frac{1}{2} \implies \cos C=\frac{1}{2}$

So, $\cos C=\frac{1}{2} \implies \cos C=\cos 60^{\circ} \implies C=\boxed{60^{\circ}}$

${ a }^{ 2 }+{ b }^{ 2 }+2*a*b{ -{ c }^{ 2 }=3*a*b }$

${ a }^{ 2 }+{ b }^{ 2 }+a*b=c^{ 2 }$

${ a }^{ 2 }+{ b }^{ 2 }-2*a*b\cos { 60={ c }^{ 2 } }$

Look (a + b + c)(a + b - c) = { (a+b) }^{ 2 }\quad -\quad { c }^{ 2 } = 3ab

{ a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad { c }^{ 2 } = ab

Cos C = \frac { { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad { c }^{ 2 } }{ 2ab }

Cos C = \frac { ab }{ 2ab }

C = 60

i have solved it just by assuming an regular triangle

so , $a = b = c$

substitute : $(a+b+c)(a+b-c) = (3a)(a) = 3a^{2} = 3ab$

Put a=b=c=1.

I solved it use y equation 1/2 abSinC with Area from Hero's formula

it must be an equilateral tianlge

take a=b=c=1 which satisfy given eq. and it must be equilateral

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

so, (a+b) ^ 2 - c ^ 2=3ab

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

Thus a^2+b^ 2-ab=a^ 2+ b ^ 2 -2abcos C

Therefore cos C = 1 / 2 =cos60°

In this way C=60°