In rectangle $A B C D$, side $A B$ is three times longer than side $B C$. Inside the rectangle, there is a point $N$, such that $A N=\sqrt{2}, B N=4 \sqrt{2}, D N=2$. Find the cosine of angle $B A N$ and the area of rectangle $A B C D$.
Denote $A D=x, \angle B A D=\alpha$, and using the cosine theorem, form a system of equations in terms of $x$ and $\alpha$.

## Solution

Let $A D=B C=x, A B=D C=3 x, \angle B A N=\alpha$. Then $\angle N A D=90^{\circ}-\alpha$. By the cosine theorem from triangles $B A N$ and $N A D$, we find that

$$
\begin{aligned}
& B N^{2}=A N^{2}+A B^{2}-2 A N \cdot A B \cos \alpha \\
& D N^{2}=A D^{2}+A N^{2}-2 A D \cdot A N \sin \alpha
\end{aligned}
$$

We obtain the system

$$
\left\{\begin{array}{l}
32=2+9 x^{2}-2 \cdot \sqrt{2} \cdot 3 x \cdot \cos \alpha \\
4=x^{2}+2-2 \cdot x \cdot \sqrt{2} \cdot \sin \alpha
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
2 x \sqrt{2} \cos \alpha=3 x^{2}-10 \\
2 x \sqrt{2} \sin \alpha=x^{2}-2
\end{array}\right.
$$

Square both sides of each equation and add the resulting equations. We get the equation

$$
8 x^{2}=9 x^{4}-60 x^{2}+100+x^{4}-4 x^{2}+4 \text {, or } 10 x^{4}-72 x^{2}+104=0 .
$$

From this, we find that $x^{2}=2$ or $x^{2}=\frac{26}{5}$. Only the second root satisfies the problem's condition (for $x=\sqrt{2}$, the triangle inequality for triangle $B A N$ is not satisfied). Therefore,

$$
S_{\mathrm{ABCD}}=3 x^{2}=\frac{78}{5}
$$

By the cosine theorem from triangle $B A N$, we find that $\cos \alpha=\frac{7}{\sqrt{65}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-26.jpg?height=366&width=876&top_left_y=1806&top_left_x=591)

## Answer

$\frac{7}{\sqrt{65}} ; \frac{78}{5}$