\(1,\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \text{Mà }\widehat{A}=\widehat{B}=\widehat{C}\\ \Rightarrow\widehat{A}=\widehat{B}=\widehat{C}=\dfrac{180^0}{3}=60^0\\ 2,\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=110^0\\ \text{Mà }\widehat{B}-\widehat{C}=10^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+10^0\right):2=60^0\\\widehat{C}=60^0-10^0=50^0\end{matrix}\right.\)

a: góc A-góc D=20 độ

góc A+góc D=180 độ

=>góc A=(20+180)/2=100 độ và góc D=180-100=80 độ

góc B=2*góc C

góc B+góc C=180 độ

=>góc B=2/3*180=120 độ; góc C=180-120=60 độ

b: góc B-góc C=20 độ

góc B+góc C=180 độ

=>góc B=(180+20)/2=100 độ và góc C=80 độ

=>góc A=100+20=120 độ

=>góc D=60 độ

a)

 

=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180^o

100^o + \(\widehat{B}+\widehat{C}\) = 180^o

\(\widehat{B}+\widehat{C}\) = 180^o - 100^o

\(\widehat{B}+\widehat{C}\) = 80^o

Góc B = (80^o+50^o):2 = 65^o

=> \(\widehat{C}\) = 65^o - 50^o = 15^o

Vậy \(\widehat{B}\) = 65^o ; \(\widehat{C}\) = 15^o

b)

 

Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180^o

\(\widehat{3A}+\widehat{2C}\) = 180^o - 80^o

\(\widehat{3A}+\widehat{2C}\) = 100^o

=> \(\widehat{A}\) = 100^o:(3+2).3 = 60^o

\(\widehat{C}\) = 100^o - 60^o = 40^o

Vậy \(\widehat{A}\) = 60^o ; \(\widehat{C}\) = 40^o

\(\widehat{D}=\dfrac{3}{2}\widehat{B}=\dfrac{3}{2}.60^0=90^0\)

\(\widehat{D}=\dfrac{4}{3}\widehat{C}\Rightarrow\widehat{C}=\dfrac{3}{4}\widehat{D}=\dfrac{3}{4}.90^0=67,5^0\)

\(\widehat{A}=360^0-\widehat{B}-\widehat{C}-\widehat{D}=360^0-60^0-90^0-67,5^0=142,5^0\)

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

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