\(\widehat{B}=180^o-\left(40^o+120^o\right)=20^o\). A C B 35 H \(AH=AB.sinB=35.sin20^o\cong12cm.\) \(\widehat{HCA}=180^o-120^o=60^o\). \(AH=AC.sin60^o\Rightarrow AC=\dfrac{AH}{sin60}=\dfrac{12}{\dfrac{\sqrt{3}}{2}}=8\sqrt{3}\). Áp dụng định lý Cô-sin: \(BC=\sqrt{AB^2+AC^2-2.AB.AC.sinA}\)\(=\sqrt{35^2+\left(8\sqrt{3}\right)^2-2.35.8\sqrt{3}.cos40^o}\cong26cm\). Vậy \(a=26cm;b=8\sqrt{3}cm,\)\(\widehat{B}=20^o\).