Mathematics - Properties of Triangle Question with Solution | TestHub

The given equation is $(a+b+c)(a+b-c)=3ab$. This can be rewritten as$((a+b)+c)((a+b)-c)=3ab$, which simplifies to $(a+b{)}^2-c^2=3ab$. Expanding $(a+b{)}^2$, we get $a^2+b^2+2ab-c^2=3ab$. Rearranging the terms, we have$a^2+b^2-c^2=ab$. From the cosine rule, $\cos C=\frac{a^2+b^2-c^2}{2ab}$. Substituting$a^2+b^2-c^2=ab$ into the cosine rule, we get $\cos C=\frac{ab}{2ab}=\frac{1}{2}$. Therefore, $C=60^{\circ }$.