[tex]\begin{gathered} v=10i-3j \\ w=-5i-j \end{gathered}[/tex]
First we need to find the dot product and the magnitude of each vector, so:
[tex]\begin{gathered} \mleft\Vert v\mright||=\sqrt[]{10^2+(-3)^2}=\sqrt[]{109} \\ \mleft\Vert w\mright||=\sqrt[]{(-5)^2+(-1)^2}=\sqrt[]{26} \end{gathered}[/tex][tex]v\cdot w=10\cdot(-5)+(-3)\cdot(-1)=-50+3=-47[/tex]
Therefore:
[tex]\begin{gathered} \theta=\cos ^{-1}(\frac{-47}{\sqrt[]{109}\cdot\sqrt[]{26}}) \\ \theta\approx152.0^{\circ} \end{gathered}[/tex]
