Answer
$90^{\circ}$
Work Step by Step
Step 1: We let $\textbf {u}=\langle 1,2 \rangle$ and $\textbf {v}=\langle -6,3 \rangle$
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: $\cos\theta=\frac{\langle 1,2 \rangle\cdot\langle -6,3 \rangle}{|\langle 1,2 \rangle||\langle -6,3 \rangle|}$
Step 4: $\cos\theta=\frac{1(-6)+2(3)}{\sqrt (1^{2}+2^{2})\cdot\sqrt ((-6)^{2}+3^{2})}$
Step 5: $\cos\theta=\frac{-6+6}{\sqrt (1+4)\cdot\sqrt (36+9)}$
Step 6: $\cos\theta=\frac{0}{\sqrt (5)\cdot\sqrt (45)}$
Step 7: $\cos\theta=0$
Step 8: $\theta=\cos^{-1}(0)$
Step 9: Solving using the inverse cos function on the calculator,
$\theta=\cos^{-1}(0)=90^{\circ}$