If point O is the outer center of triangle ABC and OA + ob + CO = 0, then what is the inner angle c,
Connecting Ao, Bo
OA=OC+CA
OA+OB+CO=0
So OC + Ca + ob + CO = 0
CA+OB=0
So | Ca | = | Bo|
Similarly, CB = Ao|
Ao | = | Bo | = | Co|
So △ AOC and △ BOC are equilateral triangles
So ∠ C = 60 ° + 60 ° = 120 °
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