Đáp án+Giải thích các bước giải:
Bài `2:`
Vì `\hat{C}=2\hat{A}`
`=>\hat{C}=2.40^0=80^0`
Mà `\hat{A}+\hat{C}+\hat{B}+\hat{D}=360^0`(Tổng các góc trong tứ giác)
`<=>40^0+80^0+\hat{B}+3\hat{B}=360^0`
`<=>4\hat{B}=360^0-120^0=240^0`
`<=>\hat{B}=\frac{240^0}{4}=60^0`
`->\hat{D}=3\hat{B}=3.60^0=180^0`
Vậy `{(\hat{C}=80^0),(\hat{B}=60^0),(\hat{D}=180^0):}`
Bài `3:`
`\hat{A}+\hat{C}=150^0`
`=>\hat{C}=150^0-\hat{A}`
Mà `\hat{B}+\hat{A}=220^0`
`=>\hat{B}=220^0-\hat{A}`
Mà `\hat{B}+\hat{C}=170^0`
`=>220^0-\hat{A}+150^0-\hat{A}=170^0`
`<=>2\hat{A}=220^0+150^0-170^0`
`<=>\hat{A}=\frac{200^0}{2}=100^0`
`=>\hat{B}=220^0-100^0=120^0`
`=>\hat{C}=150^0-100^0=50^0`
Mà `\hat{A}+\hat{B}+\hat{C}+\hat{D}=360^0`(tổng các góc trong tứ giác)
`=>\hat{D}=360^0-50^0-120^0-100^0`
`<=>\hat{D}=90^0`
Vậy `{(\hat{A}=100^0),(\hat{B}=120^0),(\hat{C}=50^0),(\hat{D}=90^0):}`
