Answer
$36.87^{\circ}$
Work Step by Step
Step 1: We let $\textbf {u}=3\textbf {i}+4\textbf {j}$ which means that $\textbf {u}=\langle 3,4 \rangle$. Also, we let $\textbf {v}=\textbf {j}$ which means that $\textbf {v}=\langle 0,1 \rangle$.
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: $\cos\theta=\frac{\langle 3,4 \rangle\cdot\langle 0,1 \rangle}{|\langle 3,4 \rangle||\langle 0,1 \rangle|}$
Step 4: $\cos\theta=\frac{3(0)+4(1)}{\sqrt (3^{2}+4^{2})\cdot\sqrt (0^{2}+1^{2})}$
Step 5: $\cos\theta=\frac{0+4}{\sqrt (9+16)\cdot\sqrt (0+1)}$
Step 6: $\cos\theta=\frac{4}{\sqrt (25)\cdot\sqrt (1)}$
Step 7: $\cos\theta=\frac{4}{5\times1}$
Step 8: $\cos\theta=\frac{4}{5}$
Step 9: $\theta=\cos^{-1}(\frac{4}{5})$
Step 10: Solving using the inverse cos function on the calculator,
$\theta=\cos^{-1}(\frac{4}{5})\approx36.87^{\circ}$
