$\begin{array}{l} \widehat C = {90^o} - \widehat B = {90^o} - {36^o} = {54^o}\\ \tan B = \dfrac{{AC}}{{AB}}\\ \Rightarrow AB = \dfrac{{AC}}{{\tan B}} = \dfrac{{10}}{{\tan {{36}^o}}} \approx 13,76\\ \sin B = \dfrac{{AC}}{{BC}} \Rightarrow BC = \dfrac{{AC}}{{\sin B}} = \dfrac{{10}}{{\sin {{36}^o}}} = 17,01 \end{array}$