Áp dụng định lý cô sin trong tam giác ABC: \(c^2=a^2+b^2-2abcosC=7^2+23^2-2.7.23.cos130\)\(\cong784cm\). Vậy \(c=28cm.\) \(cosA=\dfrac{c^2+b^2-a^2}{2bc}=\dfrac{28^2+23^2-7^2}{2.23.28}=\dfrac{158}{161}\). \(\Rightarrow\widehat{A}\cong11^o\). \(\widehat{B}=180^o-\left(\widehat{A}+\widehat{C}\right)=180^o-\left(130^o+11^o\right)=39^o\).