Đáp án:
$\widehat{x'Oy'}$ `=` $60^{o}$ ; $\widehat{x'Oy}$ `=` $120^{o}$ ; $\widehat{xOy'}$ `=` $120^{o}$
Giải thích các bước giải:
Vì $\widehat{xOy}$ đối đỉnh với $\widehat{x'Oy'}$
`=>` $\widehat{xOy}$ `=` $\widehat{x'Oy'}$ `=` $60^{o}$ ( Theo tính chất )
Ta có : $\widehat{xOy}$ `+` $\widehat{x'Oy}$ `=` $180^{o}$ ( Hai góc kề bù )
`=>` $60^{o}$ + $\widehat{x'Oy}$ `=` $180^{o}$
`=>` $\widehat{x'Oy}$ `=` $180^{o}$ `-` $60^{o}$
`=>` $\widehat{x'Oy}$ `=` $120^{o}$
Ta có : $\widehat{xOy}$ `+` $\widehat{xOy'}$ `=` $180^{o}$ ( Hai góc kề bù )
`=>` $60^{o}$ `+` $\widehat{xOy'}$ `=` $180^{o}$
`=>` $\widehat{xOy'}$ `=` $180^{o}$ `-` $60^{o}$
`=>` $\widehat{xOy'}$ `=` $120^{o}$
Vậy $\widehat{x'Oy'}$ `=` $60^{o}$ ; $\widehat{x'Oy}$ `=` $120^{o}$ ; $\widehat{xOy'}$ `=` $120^{o}$
