Áp dụng tính chất tổng 3 góc ta có
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
\(\Rightarrow\widehat{A}=180^0-\left(\widehat{B}+\widehat{C}\right)\left(1\right)\)
Theo bài ra \(\widehat{A}=180^{ }-3\widehat{C}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\widehat{B}+\widehat{C}=3\widehat{C}\)
\(\Rightarrow\widehat{B}=2\widehat{C}\)
Mà do \(\widehat{B}=70^0\Rightarrow\widehat{C}=35^0\)
Khi đó \(\widehat{A}=180^{ }-\left(\widehat{B}+\widehat{C}\right)\)\(\Rightarrow\widehat{A}=180^0-\left(35^0+70^0\right)\)
\(\Rightarrow\widehat{A}=75^0\)
Vậy ....