7. Let $O$ be the circumcenter of $\triangle A B C$. If $\overrightarrow{A O}=\overrightarrow{A B}+2 \overrightarrow{A C}$, then the value of $\sin \angle B A C$ is $\qquad$
Answer: $\frac{\sqrt{10}}{4}$.
Solution: Without loss of generality, let the circumradius $R=2$ of $\triangle ABC$. From the given conditions,
$$
2 \overrightarrow{A C}=\overrightarrow{A O}-\overrightarrow{A B}=\overrightarrow{B O} \text {, }
$$

thus $A C=\frac{1}{2} B O=1$.
Taking the midpoint $M$ of $A C$, we have $O M \perp A C$, and combining (1) we know $O M \perp B O$, and $B$ and $A$ are on the same side of line $O M$. Therefore, $\cos \angle B O C=\cos \left(90^{\circ}+\angle M O C\right)=-\sin \angle M O C=-\frac{M C}{O C}=-\frac{1}{4}$.
In $\triangle B O C$, by the cosine rule we get
$$
B C=\sqrt{O B^{2}+O C^{2}-2 O B \cdot O C \cdot \cos \angle B O C}=\sqrt{10},
$$

and in $\triangle A B C$, by the sine rule we get $\sin \angle B A C=\frac{B C}{2 R}=\frac{\sqrt{10}}{4}$.