Bài 1: 

\(\widehat{C}=180^0-120^0=60^0\)

\(\widehat{D}=360^0-130^0-90^0-60^0=80^0\)

Bài 4: 

\(\widehat{A}+\widehat{B}=360^0-80^0-70^0=210^0\)

=>\(\widehat{IAB}+\widehat{IBA}=105^0\)

hay \(\widehat{AIB}=75^0\)

Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

Mà \(\widehat{A}=\widehat{C}\), \(\widehat{B}=\widehat{D}\)

\(\Rightarrow2\widehat{A}+2\widehat{D}=360^0\)

Lại có: \(\widehat{A}=2\widehat{D}\)

\(\Rightarrow2.2\widehat{D}+2\widehat{D}=360^0\)

\(\Leftrightarrow6\widehat{D}=360^0\)

\(\Leftrightarrow\widehat{D}=60^0\)

Vậy góc B = góc D = \(60^0\)

góc A = góc C = 2 lần góc D = 2.60 = \(120^0\)

Thanks

1) Vì tổng các góc tứ giác bằng 360⁰nên

Tứ giác AEDF có \(\widehat{A}\)+\(\widehat{E}\)+\(\widehat{D}\)+\(\widehat{F}\)=360⁰

hay \(\widehat{A}\) + 35⁰+120⁰+145⁰ = 360⁰

=> \(\widehat{A}\) = 360⁰ -300⁰

Vậy \(\widehat{A}\) = 60⁰

2) Ta có \(\widehat{A}\) +\(\widehat{B}\) +\(\widehat{C}\) +\(\widehat{D}\) = 360⁰ (tổng các góc của tứ giác )

hay 100⁰+60⁰+2x+3x=360⁰

=> 5x = 200

=> x=200:5

=> x=40⁰

Vậy \(\widehat{C}\) =2.40=80⁰

\(\widehat{D}\) = 3.40=120

Ta có:

\(\widehat{A}=3\cdot\widehat{C}\\ \widehat{B}=4\cdot\widehat{C}\\ \widehat{D}=2\cdot\widehat{C}\\ \Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=3\cdot\widehat{C}+4\cdot\widehat{C}+\widehat{C}+2\cdot\widehat{C}\\ \Leftrightarrow10\cdot\widehat{C}=360^o\\ \Leftrightarrow\widehat{C}=36^o\\ \Rightarrow\widehat{A}=3\cdot\widehat{C}=3\cdot36=108^o\\ \widehat{B}=4\cdot\widehat{C}=4\cdot36^o=144^o\\ \widehat{D}=2\cdot\widehat{C}=2\cdot36^o=72^o\)

Vậy \(\widehat{A}=108^o;\widehat{B}=144^o;\widehat{C}=36^o;\widehat{D}=72^o\)

1: góc A và góc B có mối quan hệ kề một đáy

góc B và góc C có mối quan hệ trong cùng phía

2: \(\widehat{A}=180^0-40^0=140^0\)

\(\widehat{C}=180^0-80^0=100^0\)