Find the angle between $2 \mathbf{i}+5 \mathbf{j}$ and $\mathbf{j}$. The angle between $2 \mathbf{i}+5 \mathbf{j}$ and $\mathbf{j}$ is (Round to the nearest tenth as needed.)

Step 1 ：Given vectors \(\mathbf{A} = 2 \mathbf{i}+5 \mathbf{j}\) and \(\mathbf{B} = \mathbf{j}\).

Step 2 ：The dot product of \(\mathbf{A}\) and \(\mathbf{B}\) is calculated as \(\mathbf{A} \cdot \mathbf{B} = 2*0 + 5*1 = 5\).

Step 3 ：The magnitude of \(\mathbf{A}\) is calculated as \(||\mathbf{A}|| = \sqrt{2^2 + 5^2} = \sqrt{29}\).

Step 4 ：The magnitude of \(\mathbf{B}\) is calculated as \(||\mathbf{B}|| = \sqrt{0^2 + 1^2} = 1\).

Step 5 ：Using the dot product formula, we find \(\cos(\theta) = \frac{5}{\sqrt{29}*1} = \frac{5}{\sqrt{29}}\).

Step 6 ：Finally, we find the angle \(\theta\) by taking the inverse cosine of \(\cos(\theta)\), which gives us \(\theta = 21.8\) degrees.

Step 7 ：Final Answer: The angle between \(2 \mathbf{i}+5 \mathbf{j}\) and \(\mathbf{j}\) is \(\boxed{21.8}\) degrees.