Answer
$78.93^{\circ}$
Work Step by Step
Step 1: We let $\textbf {u}=-5\textbf {i}+12\textbf {j}$ which means that $\textbf {u}=\langle -5,12 \rangle$. Also, we let $\textbf {v}=3\textbf {i}+2\textbf {j}$ which means that $\textbf {v}=\langle 3,2 \rangle$.
Step 2: The formula for finding the angle between a
pair of vectors is $\cos\theta=\frac{\textbf {u}\cdot\textbf {v}}{|\textbf {u}||\textbf {v}|}$
Step 3: $\cos\theta=\frac{\langle -5,12 \rangle\cdot\langle 3,2 \rangle}{|\langle -5,12 \rangle||\langle 3,2 \rangle|}$
Step 4: $\cos\theta=\frac{-5(3)+12(2)}{\sqrt ((-5)^{2}+12^{2})\cdot\sqrt (3^{2}+2^{2})}$
Step 5: $\cos\theta=\frac{-15+24}{\sqrt (25+144)\cdot\sqrt (9+4)}$
Step 6: $\cos\theta=\frac{9}{\sqrt (169)\cdot\sqrt (13)}$
Step 7: $\cos\theta=\frac{9}{13\times\sqrt 13}$
Step 8: $\cos\theta=\frac{9}{13\sqrt 13}$
Step 9: $\theta=\cos^{-1}(\frac{9}{13\sqrt 13})$
Step 10: Solving using the inverse cos function on the calculator,
$\theta=\cos^{-1}(\frac{9}{13\sqrt 13})\approx78.93^{\circ}$
