Good afternoon!
I don't know what happened to Latex, but, now i'm posting again (fixing the problem).
Can a tutor erase the last one? With errors in Latex? Thanks!
There is another way to find the 3 angles, using a half-angle tangent formula, like this:
[tex]\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}\\\\\tan\left(\dfrac{\beta}{2}\right)=\dfrac{r}{p-b}\\\\\tan\left(\dfrac{\gamma}{2}\right)=\dfrac{r}{p-c}[/tex]
Where:
[tex]p=\dfrac{a+b+c}{2}\text{ and}[/tex]
[tex]r=\sqrt{\dfrac{(p-a)(p-b)(p-c)}{p}}[/tex]
So, to a 5:6:7 proportion, we calc p, r, and after that the 3 angles (or half-angles):
[tex]p=\dfrac{5+6+7}{2}=9[/tex]
[tex]r=\sqrt{\dfrac{(9-5)(9-6)(9-7)}{9}}=\sqrt{\dfrac{4.3.2}{9}}=\dfrac{2\sqrt{6}}{3}[/tex]
[tex]\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-5}=\dfrac{\sqrt{6}}{6}\Rightarrow \boxed{\alpha\approx 44^{\circ}\;24'}[/tex]
[tex]\tan\left(\dfrac{\beta}{2}\right)=\dfrac{r}{p-b}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-6}=\dfrac{2\sqrt{6}}{9}\Rightarrow \boxed{\beta\approx 57^{\circ}\;6'}[/tex]
[tex]\tan\left(\dfrac{\gamma}{2}\right)=\dfrac{r}{p-c}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-7}=\dfrac{\sqrt{6}}{3}\Rightarrow \boxed{\gamma\approx 78^{\circ}\;30'}[/tex]
[tex]\alpha+\beta+\gamma=180^{\circ}[/tex]
I hope I have helped!