Bài `3a)` Ta có: `\hat{AOB} + \hat{BOC} = 160^o` (gt)
`\hat{AOB} - \hat{BOC} = 120^o`. (gt)
`⇒ (\hat{AOB} + \hat{BOC}) + (\hat{AOB} - \hat{BOC}) = 160^o + 120^o = 280^o`.
`⇒ \hat{AOB} + \hat{BOC} + \hat{AOB} - \hat{BOC} = 280^o`.
`⇒ 2\hat{AOB} = 280^o`.
`⇒ \hat{AOB} = 140^o`.
`⇒ \hat{BOC} = 140^o - 120^o = 20^o`.
Vậy `\hat{AOB} = 140^o`; `\hat{BOC} = 20^o`.
`b)` Ta có: `\hat{DOB} + \hat{BOC} = \hat{DOC}`
`⇒ \hat{DOB} = \hat{DOC} - \hat{BOC} = 90^o - 20^o = 70^o`.
Ta có: `\hat{DOB} + \hat{DOA} = \hat{AOB}`
`⇒ \hat{DOA} = \hat{AOB} - \hat{DOB} = 140^o - 70^o = 70^o`.
`⇒ \hat{DOB} = \hat{DOA} = 70^o`.
`⇒ OD` là tia phân giác của `\hat{AOB}`.
`c)` Ta có: `\hat{DOC} + \hat{DOC'} = 180^o` (`2` góc kề bù)
`⇒ \hat{DOC'} = 180^o - 90^o = 90^o`.
Ta có: `\hat{DOA} + \hat{AOC'} =\hat{DOC'}`
`⇒ \hat{AOC'} = 90^o - 70^o = 20^o`.
`⇒ \hat{AOC'} = \hat{BOC} (= 20^o)`
Ta có: `\hat{AOC} = \hat{AOB} + \hat{BOC} `
`\hat{BOC'} = \hat{BOA} + \hat{AOC'}`
Mà `\hat{AOC'} = \hat{BOC}` `(cmt)`
`⇒ \hat{AOC} = \hat{BOC'}`.
Bài `4a)` Ta có: `\hat{BOC} + \hat{COA} = \hat{BOA}` (`2` góc kề bù)
`⇒ \hat{BOC} = 180^o - 50^o = 130^o`.
`b)` Ta có: `\hat{COD} + \hat{DOB} = \hat{COB}`
`⇒ \hat{COD} = 130^0 - 40^o = 90^o`.
`⇒ OC ⊥ OD`.
