Xét \(\Delta ABC.có:\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{B}+\widehat{C}=105^o\)
\(\widehat{B}=\left(105^o+25^o\right):2=65^o\)
\(\widehat{C}=105^o-65^o=40^o\)
ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow75^o+\widehat{B}+\widehat{C}=180^o\\ \Rightarrow\widehat{B}+\widehat{C}=180^o-75^o=105^o\)
do đó:\(\widehat{B}=\dfrac{\left(105^o+25^o\right)}{2}=65^o\)
\(\widehat{C}=65^o-25^o=40^o\)