Đáp án:
`\hat{A}=120^{0};\hat{B}=108^{0};\hat{C}=68^{0};\hat{D}=64^{0}`
Giải thích các bước giải:
Tứ giác `ABCD` có: `\hat{A}+\hat{B}+\hat{C}+\hat{D}=360^0`
`->6x+(5x+8)+(4x-12)+(3x+4)=360^0`
`->(6x+5x+4x+3x)+(8-12+4)=360^0`
`->18x+0=360^0`
`->18x=360^0`
`->x=360^{0}:18`
`->x=20^{0}`
Ta có: `\hat{A}=6x=6.20=120^0`
`\hat{B}=5x+8=5.20+8=108^0`
`\hat{C}=4x-12=4.20-12=68^0`
`\hat{D}=3x+4=3.20+4=64^0`
Vậy `\hat{A}=120^{0};\hat{B}=108^{0};\hat{C}=68^{0};\hat{D}=64^{0}`
