Ta có: \(\widehat B = {180^0} - \left( {\widehat A + \widehat C} \right)\)\( = {180^0} - \left( {{{40}^0} + {{120}^0}} \right) = {20^0}\)
Theo định lí sin ta có:
\(\dfrac{a}{{\sin A}} = \dfrac{c}{{\sin C}}\)\( \Rightarrow a = \dfrac{{c\sin A}}{{\sin C}} = \dfrac{{35.\sin {{40}^0}}}{{\sin {{120}^0}}} \approx 26(cm)\)
\(\dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\)\( \Rightarrow b = \dfrac{{c\sin B}}{{\sin C}} = \dfrac{{35.\sin {{20}^0}}}{{\sin {{120}^0}}} \approx 14(cm)\)
