Giải:
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{A}+\widehat{B}+30^o=180^o\)
\(\Rightarrow\widehat{A}+\widehat{B}=150^o\)
a) Ta có: \(\widehat{A}+\widehat{B}=150^o\)
\(\Rightarrow x+y=150^o\)
Mà x = 2y
\(\Rightarrow2y+y=150^o\)
\(\Rightarrow3y=150^o\)
\(\Rightarrow y=50^o\)
\(\Rightarrow x=50^o.2=100^o\)
Vậy \(y=50^o,x=100^o\)
b) Ta có: \(\widehat{A}+\widehat{B}=150^o\)
\(\Rightarrow x+y=150^o\)
Mà \(x-y=10^o\)
\(\Rightarrow x=\left(150^o+10^o\right):2=80^o\)
\(\Rightarrow y=150^o-80^o=70^o\)
Vậy \(x=80^o,y=70^o\)
c) Ta có: \(3x=2y\Rightarrow\frac{x}{3}=\frac{y}{2}\) và \(\widehat{A}+\widehat{B}=150^o\) hay \(x+y=150^o\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{2}=\frac{x+y}{3+2}=\frac{150^o}{5}=30^o\)
+) \(\frac{x}{3}=30^o\Rightarrow x=90^o\)
+) \(\frac{y}{2}=30^o\Rightarrow y=60^o\)
Vậy \(x=90^o,y=60^o\)