Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}=180^0-30^0-45^0=105^0\)
Xét ΔABC có \(\frac{AB}{\sin C}=\frac{AC}{\sin B}=\frac{BC}{\sin A}\)
=>\(\frac{AB}{\sin45}=\frac{BC}{\sin105}=\frac{4}{\sin30}=8\)
=>\(AB=8\cdot\sin45\) ≃6(cm); BC=8*sin105≃8(cm)

