Explanation
Step 1
let
[tex]\begin{gathered} v=\langle-3,-2\rangle \\ w=\langle1,5\rangle \end{gathered}[/tex]we can use the product scalar formula
[tex]\begin{gathered} \vec{v}\cdot\vec{w}=\lvert\vec{v}\rvert\lvert\vec{w}\rvert\cos \emptyset \\ \end{gathered}[/tex]where theta, is the angle between the vectors, so
a) find the measure of the vector v
[tex]\begin{gathered} \lvert\vec{v}\rvert=\sqrt[]{(-3)^2+(-2)^2} \\ \lvert\vec{v}\rvert=\sqrt[]{9+4} \\ \lvert\vec{v}\rvert=\sqrt{13} \end{gathered}[/tex]b) now the other vector (w)
[tex]\begin{gathered} \lvert\vec{w}\rvert=\sqrt[]{(1)^2+(5)^2} \\ \lvert\vec{w}\rvert=\sqrt[]{1+25} \\ \lvert\vec{w}\rvert=\sqrt[]{26} \end{gathered}[/tex]Step 2
find the scalar product.
The angle between two vectors can be found using vector multiplication (scalar product)
[tex]\begin{gathered} \text{scalar product} \\ a=\langle x_1,y_1\rangle \\ b=\langle x_2,y_2\rangle \\ a\cdot b=(x_1\cdot x_2+y_1\cdot y_2) \end{gathered}[/tex][tex]\vec{v}\cdot\vec{w}=(-3\cdot1-2\cdot5)=-3-10=-13[/tex]Step 3
Finally, replace in the formula
[tex]\begin{gathered} \vec{v}\cdot\vec{w}=\lvert\vec{v}\rvert\lvert\vec{w}\rvert\cos \emptyset \\ -13=\sqrt[]{13}\sqrt[]{26}\cos \text{ }\emptyset \\ -13=18.38\text{ cos}\emptyset \\ \text{divide both sides by 18.38} \\ \frac{-13}{18.38}=\frac{18.38\text{ cos}\emptyset}{18.38} \\ -0.07054=\text{ cos }\emptyset \\ \emptyset\text{= }\cos ^{-1}(-0.070) \\ \emptyset=135.01 \\ \text{rounded} \\ \emptyset=135 \end{gathered}[/tex]I hope this helps you
