[tex]\it \Delta A'OB-dreptunghic,\ \widehat{O}=90^o, \stackrel{T.P.}{\Longrightarrow}\ BO^2=A'B^2-A'O^2=4^2-(\sqrt2)^2=\\ \\ =16-2=14 \Rightarrow BO=\sqrt{14}\ cm\\ \\ tg(BA'O)=\dfrac{BO}{A'O}=\dfrac{\sqrt{14}}{\sqrt2}=\sqrt{\dfrac{14}{2}}=\sqrt7[/tex]
