Do ABC vuông tại A nên \(\widehat{B}+\widehat{C}=90^0\)
\(\Rightarrow3\widehat{B}+3\widehat{C}=270^0\)
\(\Rightarrow\widehat{B}+\left(2\widehat{B}+3\widehat{C}\right)=270^0\)
\(\Rightarrow\widehat{B}+220^0=270^0\)
\(\Rightarrow\widehat{B}=50^0\)
\(\Rightarrow\widehat{C}=90^0-\widehat{B}=40^0\)