Solution;
[tex]\begin{gathered} v=5i-3j \\ w=2i-j \end{gathered}[/tex][tex]\begin{gathered} v.w=\lvert{v}\rvert\lvert{w}\rvert cos\theta \\ cos\theta=\frac{v.w}{\lvert{v}\rvert\lvert{w}\rvert} \end{gathered}[/tex][tex]\begin{gathered} v.w=5(2)+(-3)(-1)=10+3=13 \\ \lvert{v}\rvert=\sqrt{5^2+(-3)^2}=\sqrt{34} \\ \lvert{w}\rvert=\sqrt{2^2+(-1)^2}=\sqrt{5} \\ \end{gathered}[/tex][tex]\begin{gathered} cos\theta=\frac{13}{(\sqrt{5)(\sqrt{34})}}=\frac{13}{13.03840481}=0.9971 \\ \theta=\cos^{-1}(0.9971) \\ \theta=4.40\degree \end{gathered}[/tex]The answer is 4.40°
