Xét ΔACB có \(\widehat{A}+\widehat{C}+\widehat{B}=180^0\)
=>\(\widehat{B}=180^0-30^0-45^0=105^0\)
Xét ΔABC có \(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\)
=>\(\dfrac{8}{sin30}=\dfrac{AC}{sin105}=\dfrac{AB}{sin45}\)
=>\(\dfrac{AC}{sin105}=\dfrac{BC}{sin45}=16\)
=>\(\left\{{}\begin{matrix}AC=16\cdot sin105=4\sqrt{6}+4\sqrt{2}\\BC=16\cdot sin45=8\sqrt{2}\end{matrix}\right.\)