Xét $\Delta{ABC}$:
$\widehat{A}+\widehat{B}+\widehat{C}=180^o$
mà $\widehat{B}=\dfrac{7}{6}\widehat{C},\widehat{A}=\dfrac{5}{6}\widehat{C}$
$\to \dfrac{5}{6}\widehat{C}+\dfrac{7}{6}\widehat{C}+\widehat{C}=180^o$
$\to \widehat{C}\Big(\dfrac{5}{6}+\dfrac{7}{6}+1\Big)=\widehat{C}.3=180^o$
$\to \widehat{C}=60^o$
$\to \begin{cases}\widehat{A}=50^o\\\widehat{B}=70^o\end{cases}$