\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=80^0\\ \text{Mà }\widehat{B}-\widehat{C}=40^0\\ \Rightarrow\widehat{B}=\left(80^0+40^0\right):2=60^0\\ \Rightarrow\widehat{C}=60^0-40^0=20^0\)
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\(\widehat{B}=60^0\)
\(\widehat{C}=20^0\)
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