Ta có: $AB = \dfrac{1}{2}AC$
⇒ $BC = \sqrt{AB^{2} + AC^{2}} = \sqrt{(\dfrac{AC}{2})^{2} + AC^{2}} = \sqrt{\dfrac{5AC^{2}}{4}} = \dfrac{AC\sqrt{5}}{2}$
$\begin{cases}sin\widehat{C} = \dfrac{AB}{BC} = \dfrac{\dfrac{AC}{2}}{\dfrac{AC\sqrt{5}}{2}} = \dfrac{\sqrt{5}}{5}\\cos\widehat{C} = \dfrac{AC}{BC} = \dfrac{AC}{\dfrac{AC\sqrt{5}}{2}}=\dfrac{2\sqrt{5}}{5}\\tan\widehat{C} = \dfrac{sin\widehat{C}}{cos\widehat{C}}=\dfrac{\dfrac{\sqrt{5}}{5}}{\dfrac{2\sqrt{5}}{5}} = \dfrac{1}{2}\\cot\widehat{C} = \dfrac{1}{tan\widehat{C}} = 2\end{cases}$
⇒ $\begin{cases}sin\widehat{B} = cos\widehat{C} = \dfrac{2\sqrt{5}}{5}\\cos\widehat{B} = sin\widehat{C} = \dfrac{\sqrt{5}}{5}\\tan\widehat{B} = cot\widehat{C} = 2\\cot\widehat{B} = tan\widehat{C} = \dfrac{1}{2} \end{cases}$
