Đáp án: `\hat{A}=140^o;\hat{B}=90^o`
Giải thích các bước giải:
Vì `DI` là p/g `\hat{ADC}`
`=>\hat{IDC}=1/2\hat{ADC}`
`=>\hat{ADC}=2\hat{IDC}`
C/m tương tự : `=>\hat{BCD}=2\hat{ICD`
Xét `ΔIDC` có : `\hat{IDC}+\hat{ICD}+115^o=180^o`
`=>\hat{IDC}+\hat{ICD}=65^o`
`=>2\hat{IDC}+2\hat{ICD}=130^o`
Vì `\hat{A}-\hat{B}=50^o`
`=>\hat{A}=50^o+\hat{B`
Xét tứ giác `ABCD` có :
`\hat{A}+\hat{B}+\hat{ADC}+\hat{BCD}=360^o`
`=>50^o+\hat{B}+\hat{B}+2\hat{ICD}+2\hat{IDC}=360^o`
`=>50^o+2\hat{B}+130^o=360^o`
`=>2\hat{B}=180^o`
`=>\hat{B}=90^o`
`=>\hat{A}=90^o +50^o=140^o`
Vậy `\hat{A}=140^o;\hat{B}=90^o`
