[tex]of \ pentacontagon \ has \ 50 \ interior \ angles \\ The \ sum \ of \ the \ interior \ angles \ of \ a \ polygon \ can \ be \ found \ using \ the \ formula:\\ (n-2) \cdot 180^o \\n=50\\\\( n-2) \cdot 180^o=(50-2) \cdot 180^o=48\cdot 180^o = 8\ 640^o[/tex]
