\(\dfrac{A}{1}=\dfrac{B}{2}=\dfrac{C}{3}=\dfrac{D}{4}=\dfrac{A+B+C+D}{1+2+3+4}=\dfrac{360}{10}=36\)

\(\Rightarrow A=36^0;B=36.2=72^0;C=36.3=108^0;D=36.4=144^0\)

Trong tứ giác \(ABCD\), tổng các góc bằng \(360^\circ \) nên ta có:

\(\begin{array}{l}x + 2x + 3x + 4x = 360^\circ \\10x = 360^\circ \\x = 360^\circ :10\\x = 36^\circ \end{array}\)

Suy ra:

\(\widehat A = 36^\circ ;\;\widehat B = 72^\circ ;\;\widehat C = 108^\circ ;\;\widehat D = 144^\circ \)

góc C-góc D=200-180=20 độ

góc C+góc D=120 độ

=>góc C=(20+120)/2=70 độ và góc D=120-70=50 độ

góc B=200-70=130 độ

góc A=180-70=110 độ

TA CÓ : 

+ \(\widehat{C}=2\widehat{D}\) \(\Rightarrow\widehat{D}=\frac{1}{2}\widehat{C}\)

+ \(\widehat{B}=2\widehat{C}\)

=>  \(\widehat{A}=\widehat{B}+\widehat{D}=2\widehat{C}+\frac{1}{2}\widehat{C}=\frac{5}{2}.\widehat{C}\)

Mặt khác  vì ABCD là 1 tứ giác nên  \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

hay  \(\frac{5}{2}.\widehat{C}+2\widehat{C}+\widehat{C}+\frac{1}{2}.\widehat{C}=6\widehat{C}=360^o\)

=>  \(\widehat{C}=60^o\)=> \(\widehat{B}=2.\widehat{C}=60^o.2=120^o\) ;  \(\widehat{A}=\frac{5}{2}.\widehat{C}=\frac{5}{2}.60^o=150^o\);\(\widehat{D}=\frac{1}{2}.60^o=30^o\)

góc C-góc D=10

=>góc C=góc D+10

góc B-góc C=10

=>góc B=10+góc C=góc D+20

góc A-góc B=10

=>góc A=góc B+10=góc D+30

góc A+góc B+góc C+góc D=360

=>4*góc D+60=360

=>góc D=75 độ

=>góc C=85 độ; góc B=95 độ; góc A=105 độ