\(\widehat{A}=63^o\Rightarrow\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-63^o=117^o\\ \Rightarrow\left\{{}\begin{matrix}\widehat{B}=117^o-\widehat{C}\\\widehat{C}=117^o-\widehat{B}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\widehat{MBC}=\dfrac{117^o-\widehat{C}}{2}\\\widehat{MCB}=\dfrac{117^o-\widehat{B}}{2}\end{matrix}\right.\\ \Rightarrow\widehat{MBC}+\widehat{MCB}=\dfrac{117^o-\widehat{C}}{2}+\dfrac{117^o-\widehat{B}}{2}\\ =\dfrac{117^o-\widehat{C}+117^o-\widehat{B}}{2}\\ =\dfrac{117^o+117^o-\left(\widehat{B}+\widehat{C}\right)}{2}=\dfrac{117^o+117^o-117^o}{2}\\ =\dfrac{117^o}{2}=58,5^o\\ \widehat{BMC}=180^o-\left(\widehat{MBC}+\widehat{MCB}\right)=180^o-58,5^o=121,5^o\)
Vậy \(\widehat{BMC}=121,5^o\)
