Chapter 12 - Vectors and Matrices - 12.2 The Components of a Vector - Exercises and Problems for Section 12.2 - Exercises and Problems - Page 509: 15

$\dfrac{\pi}{4}$

We see that the initial point is: $(1,5)$. The ending point is: $(2,4)$. So, $\vec{u} \cdot \vec{v}=(2-1)\vec{i}+(4-5) \vec{j}=\vec{i}-\vec{j}$ Now, the angle between two vectors can be computed as: $\cos \theta =\dfrac{\vec{u} \cdot \vec{v}}{|u||v|}=\dfrac{(\vec{i}-\vec{j}) (-\vec{j})}{\sqrt {1+1}}\\=\dfrac{1}{\sqrt 2}$ Therefore, $\theta =\cos^{-1}(\dfrac{1}{\sqrt 2})=\dfrac{\pi}{4}$