Đáp án:
Ta có
`\hat{BOC} + \hat{COD} + \hat{DOA} = \hat{AOB}`
` => \hat{BOC} + \hat{COD} + \hat{DOA} = 160^0`
` => (\hat{BOD} - \hat{COD}) + \hat{COD} + (\hat{AOC} - \hat{COD}) = 160^0`
` => 90^0 - \hat{COD} + \hat{COD} + 90^0 - \hat{COD} = 160^0`
` => 180^0 - \hat{COD} = 160^0`
` => \hat{COD} = 20^0`