(4) Given that $O$ is the circumcenter of acute $\triangle A B C$, $A B=6, A C=10$, if $\overrightarrow{A O}=$ $x \overrightarrow{A B}+y \overrightarrow{A C}$, and $2 x+10 y=5$, then $\cos \angle B A C=$ $\qquad$ .
(4) $\frac{1}{3}$ Hint: Since $O$ is the circumcenter of $\triangle A B C$, we have
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{A C}=|\overrightarrow{A O}| \cdot|\overrightarrow{A C}| \cdot \cos \angle O A C=\frac{1}{2}|\overrightarrow{A C}|^{2}=50 . \\
\text { Also, } \overrightarrow{A O}=x \overrightarrow{A B}+y \overrightarrow{A C} \text {, so } \\
\quad(x \overrightarrow{A B}+y \overrightarrow{A C}) \cdot \overrightarrow{A C}=50,
\end{array}
$$

which means
$$
x \cdot|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos \angle B A C+y \cdot|\overrightarrow{A C}|^{2}=50 \text {, }
$$

or
$$
60 x \cdot \cos \angle B A C+100 y=50 .
$$

Since $\triangle A B C$ is an acute triangle, it is easy to see that $x \neq 0$, so
$$
\cos \angle B A C=\frac{5-10 y}{6 x}=\frac{2 x}{6 x}=\frac{1}{3} .
$$