Given the vectors $\overrightarrow{u}=7i+10j$ and $\overrightarrow{v}=3i+2j$, find the angle (in degrees) between them.

Step 1 ：Given the vectors $\overrightarrow{u}=7i+10j$ and $\overrightarrow{v}=3i+2j$, we are to find the angle (in degrees) between them.

Step 2 ：The angle between two vectors can be found using the dot product formula: $$\overrightarrow{u}\cdot \overrightarrow{v}=||\overrightarrow{u}||||\overrightarrow{v}||\cos (\theta )$$ where $\overrightarrow{u}\cdot \overrightarrow{v}$ is the dot product of $\overrightarrow{u}$ and $\overrightarrow{v}$, $||\overrightarrow{u}||$ and $||\overrightarrow{v}||$ are the magnitudes of $\overrightarrow{u}$ and $\overrightarrow{v}$ respectively, and $\theta$ is the angle between $\overrightarrow{u}$ and $\overrightarrow{v}$.

Step 3 ：We can rearrange this formula to solve for $\theta$: $$\theta ={\cos }^{-1}\left(\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{||\overrightarrow{u}||||\overrightarrow{v}||}\right)$$

Step 4 ：We can calculate the dot product $\overrightarrow{u}\cdot \overrightarrow{v}$ as $u_i\cdot v_i+u_j\cdot v_j$, and the magnitudes $||\overrightarrow{u}||$ and $||\overrightarrow{v}||$ as $\sqrt{u_i^2+u_j^2}$ and $\sqrt{v_i^2+v_j^2}$ respectively.

Step 5 ：Let's calculate these values and find the angle.

Step 6 ：For the given vectors, the dot product is 41, the magnitude of $\overrightarrow{u}$ is approximately 12.21, and the magnitude of $\overrightarrow{v}$ is approximately 3.61.

Step 7 ：Substituting these values into the formula for $\theta$, we find that the angle in radians is approximately 0.37.

Step 8 ：Converting this angle to degrees, we find that the angle between the vectors $\overrightarrow{u}=7i+10j$ and $\overrightarrow{v}=3i+2j$ is approximately $\boxed{{\displaystyle 21.32}}$ degrees.