Giải:

`a)``-` Hai góc kề bù:

`+``\hat{yOm}` và `\hat{xOy}`

`b)` Có: `\hat{yOm}``+``\hat{xOy}``=``\hat{xOm}` `(`kề bù`)`

`=>``\hat{yOm}``+``60^o``=``180^o`

`=>``180^o``-``60^o``=``\hat{yOm}`

`=>``120^o``=``\hat{yOm}`

Vậy `\hat{yOm}``=``120^o`

`c)` Vì `Ot` là tia phân giác `\hat{xOy}` `(g t)`

`=>``\hat{tOy}``=``\hat{tOx}``=``60^o : 2``=``30^o`

Vậy `\hat{tOy}``=``30^o`

Lại có: `\hat{tOm}``+``\hat{tOx}``=``\hat{xOm}`

`=>``\hat{tOm}``+``30^o``=``180^o`

`=>``180^o``-``30^o``=``\hat{tOm}`

`=>``150^o``=``\hat{tOm}`

Vậy `\hat{tOm}``=``150^o`