Find the angle $\theta$ (in degrees) between the vectors. (Round your answer to two decimal places.)
$$\begin{array}{l}\mathbf{u}=3\mathbf{i}-6\mathbf{j}\\ \mathbf{v}=9\mathbf{i}+3\mathbf{j}\end{array}$$
$$\theta =$$

Step 1 ：We are given the vectors $\mathbf{u}=3\mathbf{i}-6\mathbf{j}$ and $\mathbf{v}=9\mathbf{i}+3\mathbf{j}$.

Step 2 ：We can calculate the dot product of the vectors using the formula $\mathbf{u}\cdot \mathbf{v}=u_x{v}_x+u_y{v}_y$, where $u_x$ and $u_y$ are the x and y components of $\mathbf{u}$, and $v_x$ and $v_y$ are the x and y components of $\mathbf{v}$. Plugging in the given components, we get $\mathbf{u}\cdot \mathbf{v}=9$.

Step 3 ：We can calculate the magnitudes of the vectors using the formula $||\mathbf{u}||=\sqrt{u_x^2+u_y^2}$ and $||\mathbf{v}||=\sqrt{v_x^2+v_y^2}$. Plugging in the given components, we get $||\mathbf{u}||=6.708203932499369$ and $||\mathbf{v}||=9.486832980505138$.

Step 4 ：We can find the angle between the vectors using the formula $\theta ={\cos }^{-1}\left(\frac{\mathbf{u}\cdot \mathbf{v}}{||\mathbf{u}||||\mathbf{v}||}\right)$. Plugging in the values we calculated, we get ${\theta }_{rad}=1.4288992721907328$.

Step 5 ：We convert the angle from radians to degrees to get ${\theta }_{deg}=81.87$.

Step 6 ：Final Answer: The angle $\theta$ between the vectors $\mathbf{u}$ and $\mathbf{v}$ is $\boxed{{\displaystyle 81.87}}$ degrees.