\(\Delta ABC\) has \(\widehat{A}=60^o\Rightarrow\widehat{B}+\widehat{C}=120^o\)

and \(\dfrac{\widehat{B}}{3}=\widehat{C}\Rightarrow\widehat{B}=3\cdot\widehat{C}\Rightarrow\widehat{B}+\widehat{C}=3\cdot\widehat{C}+\widehat{C}=4\cdot\widehat{C}=120^o\)

\(\Rightarrow\widehat{C}=\dfrac{120^o}{4}=30^o\Leftrightarrow\widehat{B}=120^o-30^o=90^o\)

So \(\left\{{}\begin{matrix}\widehat{B}=90^o\\\widehat{C}=30^o\end{matrix}\right.\)