\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{18^2+20^2-14^2}{2.18.20}=\dfrac{11}{15}\).
Vậy \(\widehat{A}=42^o50'\).
\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{14^2+20^2-18^2}{2.14.20}=\dfrac{17}{20}\).
Vậy \(\widehat{B}=60^o56'\).
Vậy \(\widehat{C}=180^o-\widehat{A}-\widehat{B}=77^o46'\).