Xét tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)( tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-70^0=110^0\)
Xét tam giác ABC có:
\(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=110^0\\\widehat{B}-\widehat{C}=40^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+40^0\right):2=75^0\\\widehat{C}=\left(110^0-40^0\right):2=35^0\end{matrix}\right.\)
Ta có: \(\widehat{C}< \widehat{A}< \widehat{B}< 90^0\)
Vậy tam giác ABC là tam giác nhọn