Xét \(\Delta ABC\) có : 

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(Định lí tổng 3 góc của 1\(\Delta\))

\(50^0+\widehat{B}+\widehat{C}=180^0\)

\(\widehat{B}+\widehat{C}=180^0-50^0\)

\(\widehat{B}+\widehat{C}=130^0\)

Theo bài ra ta có : 

\(\widehat{B}:\widehat{C}=2:3\)

\(\Rightarrow\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}\) và \(\widehat{B}+\widehat{C}=130^0\)

Áp dụng tính chất của dãy tỉ số bằng nhau : 

\(\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{B}+\widehat{C}}{2+3}=\dfrac{130^0}{5}=26^0\)

\(\rightarrow\dfrac{\widehat{B}}{2}=26^0\Rightarrow\widehat{B}=26^0.2=52^0\)

\(\rightarrow\dfrac{\widehat{C}}{3}=26^0\Rightarrow\widehat{C}=26^0.3=78^0\)

Theo bài ra ta có : 

\(\widehat{A}:\widehat{B}:\widehat{C}=1:2:3\)

\(\Rightarrow\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}\) và \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

Áp dụng tính chất của dãy tỉ số bằng nhau : 

\(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{1+2+3}=\dfrac{180^0}{6}=30^0\)

\(\rightarrow\dfrac{\widehat{A}}{1}=30^0\Rightarrow\widehat{A}=30^0.1=30^0\)

\(\rightarrow\dfrac{\widehat{B}}{2}=30^0\Rightarrow\widehat{B}=30^0.2=60^0\)

\(\rightarrow\dfrac{\widehat{C}}{3}=30^0\Rightarrow\widehat{C}=30.3=90^0\)

1c

2a

3b

4b

5a

6a