`Hình 1`
` Ta có:`
`=> hat\{B}=180^0-hat\{BAC}-hat\{C}`
`=> hat\{B}=180^0-110^0-25^0`
`=>hat\{B} =45^0`
`Vậy hat\{x}=45^0`
`Ta có: hat\{y}+hat\{BAC}=180^0( hai góc kề bù)`
`=>hat\{y}=180^0-hat\{BAC}`
`=>hat\{y}=180^0-110^0`
`=>hat\{y}=70^0`
`Hình 2`
`Ta có:hat\{NPM}+hat\{MPQ}=180^0( hai góc kề bù)`
`hat\{NPM}=180^0-hat\{MPQ}`
`hat\{NPM}=180^0-110^0`
`hat\{NPM}=70^0`
`Vì triangle MNP=hat\{NPM}+hat\{N}+hat\{NMP}=180^0`
`=>hat\{NMP}=180^0-55^0-70^0=55^0`
`=>hat\{MQP}=55^0`
`Hay x=55^0`
`Vì triangle MPQ =180^0=hat\{MQP}+hat\{MPQ}+hat\{QMP}`
`=>hat\{QMP}=180^0-110^0-55^0=15^0`
`Hay y=15^0`
Chúc bạn học tốt:33
