`Ta` `có:` `\triangleABC=\triangleDEF`
`=>{(\hat{A}=\hat{D}=70^o),(\hat{B}=\hat{E}=50^o),(\hat{C}=\hat{F}):}(\text{2 góc tương ứng})`
`Xét` `\triangleABC` `có:`
`\hat{A}+\hat{B}+\hat{C}=180^o(\text{Tổng 3 góc trong tam giác})`
`=>70+50+c=180^o`
`=>\hat{C}=180^o-70^o-50^o`
`=>\hat{C}=60^o`
`=>\hat{C}=\hat{F}=60^o`
`Vậy` `\hat{A}=70^o; \hat{B}=50^o; \hat{E}=60^o; \hat{D}=70^o; \hat{E}=50^o; \hat{F}=60^o`