Có ^A + ^B =180^o
mà ^A-^B=20^o
=> ^A=(180^o+20^o):2=100^o
=> ^B=180^o-100^o=80^o
Ta có ^A+^C =150^o
=> 100+^C =150^o
=> ^C =50^o
Ta có ^C + ^D =180^o
=> ^D =180^o - ^C =180^o-50^o=130^o
Vậy ^A=100^o, ^B =80^o, ^C =50^o, ^D =130^o

Ta có :  ^A + ^B =180^o
mà ^A-^B=20^o
=> ^A=(180^o+20^o):2=100^o
=> ^B=180^o-100^o=80^o
Ta có ^A+^C =150^o
=> 100+^C =150^o
=> ^C =50^o
Ta có ^C + ^D =180^o
=> ^D =180^o - ^C =180^o-50^o=130^o
Vậy ^A=100^o, ^B =80^o, ^C =50^o, ^D =130^o

#))Giải :

Ta có :

\(\widehat{A}+\widehat{B}=180^o\)(cặp góc trong cùng phía bù nhau)

\(\widehat{A}-\widehat{B}=20^o\)

\(\Rightarrow\widehat{A}=\left(180^o+20^o\right)\div2=100^o\)

\(\Rightarrow\widehat{B}=\left(180^o-20^o\right)\div2=80^o\)

\(\widehat{A}+\widehat{C}=150^o\)

\(\Rightarrow\widehat{C}=150^o-\widehat{A}=150^o-100^o=50^o\)

\(\widehat{C}+\widehat{D}=180^o\)(cặp góc trong cùng phía bù nhau)

\(\Rightarrow\widehat{D}=180^o-\widehat{C}=180^o-50^o=130^o\)

Vậy \(\widehat{A}=100^o;\widehat{B}=80^o;\widehat{C}=50^o;\widehat{D}=130^o\)

ban chua cho biet day

Theo đề ta có: A-B= 20° => B=A-20 ; A-C=150° => C=A-150

Do ABCD là hình thang nên:

A+D=180°

B+C=180°<=>( A-20°)+(A-150°)=180° <=> 2A-170°=180°<=>2A=350°<=>A=175°

Thay A=175 vào: A-B=20° <=> 175°-B=20°<=> B=155°

A-C=150°<=> 175°-C=150° <=> C= 25°

Vậy A=175°; B=155°; C= 25°; D=5°

 Ta có: AB// CD => \(\widehat{A}\) +\(\widehat{D}\)=\(180^0\)(trong cùng phía)

Mà \(\widehat{A}\)-\(\widehat{D}\)= \(20^0\) => \(\widehat{A}\) = (\(180^0\)-\(20^0\)):2 = 100; \(\widehat{D}\)=\(80^0\) 

tương tự \(\widehat{B}\)+ \(\widehat{C}\)= \(180^0\)(trong cùng phía);\(\widehat{B}\) = 2.\(\widehat{C}\) => 3.\(\widehat{C}\) = 180 

=> \(\widehat{C}\) =60; \(\widehat{B}\) = 120