A soma dos ângulos internos de um polígono é dada pela fórmula:
[tex]\Large\displaystyle\text{$\begin{gathered} \sf S_i = (n - 2) \cdot180 {}^{ \circ} \end{gathered}$}[/tex]
Onde:
[tex]\Large\displaystyle\text{$\begin{gathered} \begin{cases} \sf S_i = soma \,dos\, \hat{a}ngulos \, internos=? \\ \sf n = n\acute{u}mero \,de\, lados = 8\end{cases}\end{gathered}$}[/tex]
Calculando a soma dos ângulos internos de um octógono pela fórmula temos que:
[tex]\Large\displaystyle\text{$\begin{gathered} \sf S_i = (n - 2) \cdot180 {}^{ \circ} \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \sf S_i = (8 -2) \cdot180 {}^{ \circ} \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \sf S_i = 6 \cdot180 {}^{ \circ} \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \sf S_i = 1080 {}^{ \circ} \end{gathered}$}[/tex]
Portanto, a soma dos ângulos internos de um octógono é:
[tex]\Large\displaystyle\text{$\begin{gathered} \boxed{ \boxed{\bf 1080 {}^{ \circ} }} \end{gathered}$}[/tex]
