4.26. In square $A B C D$, points $M$ and $N$ are the midpoints of sides $D C$ and $B C$. Find $\angle M A N$.
4.26. Let's introduce a coordinate system with the origin at point $A(0 ; 0)$ and coordinate axes directed along the sides $A D$ and $A B$ of the square (Fig. 4.23). In this system, the vector $\overline{A M}$ has coordinates $(0.5 ; 1)$, and the vector $\overline{A N}$ has coordinates $(1 ; 0.5)$. We will use the formula

$$
\begin{aligned}
& \cos \alpha=\frac{\overline{A M} \cdot \overline{A N}}{|\overline{A M}||\overline{A N}|}, \text { where } \overline{A M} \cdot \overline{A N}=0.5 \cdot 1+1 \cdot 0.5=1 \\
& |\overline{A M}|=\sqrt{0.5^{2}+1^{2}}=0.5 \sqrt{5}, \overline{A N}=\sqrt{1^{2}+0.5^{2}}=0.5 \sqrt{5}
\end{aligned}
$$

Thus, $\cos \alpha=\frac{1}{0.5 \sqrt{5} \cdot 0.5 \sqrt{5}}=\frac{4}{5}$.

$a$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-585.jpg?height=236&width=527&top_left_y=1209&top_left_x=46)

Fig. 4.24
Answer: $\arccos 0.8$.