Find the segments $B N$ and $N C$ and apply the Law of Cosines.

## Solution

From the right triangles $A N C$ and $A N B$, we find that

$$
C N=\sqrt{A C^{2}-A N^{2}}=\sqrt{0.76}, \quad B N=\sqrt{A B^{2}-A N^{2}}=2.4
$$

Therefore, $B C=2.4+\sqrt{0.76}$.

By the Law of Cosines in triangle $A B C$, we find that

$$
\begin{gathered}
\cos \angle B A C=\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C}=\frac{13-(2.4+\sqrt{0.76})^{2}}{12}= \\
=0.52-0.4 \sqrt{0.76}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-26.jpg?height=434&width=640&top_left_y=0&top_left_x=709)

## Answer

$\arccos (0.52-0.4 \sqrt{0.76})$.
[ Diameter, main properties ]

In triangle $ABC$, a circle is described on side $AC$ as a diameter, which intersects side $AB$ at point $M$ and side $BC$ at point $N$. It is known that $AC=2$, $AB=3$, $AN=1.8$. Find the cosine of angle $BAC$.