\(\widehat{C}=180^0-\widehat{A}-\widehat{B}=105^0\)
Theo định lý hàm sin:
\(\frac{a}{sinA}=\frac{c}{sinC}\Rightarrow a=\frac{c.sinA}{sinC}=\frac{4.sin30^0}{sin105^0}=2\left(\sqrt{6}-\sqrt{2}\right)\)
Diện tích tam giác:
\(S=\frac{1}{2}ac.sinB=\frac{1}{2}4.2\left(\sqrt{6}-\sqrt{2}\right).sin45^0=2,93\left(cm^2\right)\)