[tex]\it \widehat C=\widehat{DAB}\ (au\ acela\c{s}i\ complement, \widehat B)\\ \\ c)\ cosC=cos(DAB)=\dfrac{AD}{AB}=\dfrac{\sqrt3}{3\sqrt3}=\dfrac{1}{3}\\ \\ \\ \Delta DAB-dr,\ \widehat D=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ BD^2=AB^2-AD^2=(3\sqrt3^2)-(\sqrt3)^2=\\ \\ \\ =(3\sqrt3-\sqrt3)(3\sqrt3+\sqrt3)=2\sqrt3\cdot4\sqrt3=2\cdot4\cdot3=4\cdot6\\ \\ \\ BD=\sqrt{4\cdot6}=2\sqrt6\\ \\ \\ tgC=tg(DAB)=\dfrac{BD}{AD}=\dfrac{2\sqrt6}{\sqrt3}=\dfrac{2\sqrt2\sqrt3}{\sqrt3}=2\sqrt2[/tex]
