B. Let the circumcenter and orthocenter of $\triangle A B C$ be $O$ and $H$ respectively. If $B, C, H, O$ are concyclic, find all possible measures of $\angle B A C$ for all $\triangle A B C$.

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B. Discuss in three cases.
(1) $\triangle A B C$ is an acute triangle (as shown in Figure 8).

From $\angle B H C=180^{\circ}-\angle A, \angle B O C=2 \angle A$,
. Then $\angle B H C=\angle B O C$
$$
\begin{array}{l}
\Rightarrow 180^{\circ}-\angle A=2 \angle A \\
\Rightarrow \angle A=60^{\circ} .
\end{array}
$$
(2) $\triangle A B C$ is an obtuse triangle (as shown in Figure 9).

When $\angle A>90^{\circ}$, from
$$
\begin{array}{l}
\angle B H C=180^{\circ}-\angle A, \\
\angle B O C=2\left(180^{\circ}-\angle A\right), \\
\text { then } \angle B H C+\angle B O C=180^{\circ} \\
\Rightarrow 3\left(180^{\circ}-\angle A\right)=180^{\circ} \\
\Rightarrow \angle A=120^{\circ} ;
\end{array}
$$

When $\angle A<90^{\circ}$, the same as (1) we get $\angle A=60^{\circ}$.
(3) $\triangle A B C$ is a right triangle.

When $\angle A=90^{\circ}$, since $O$ is the midpoint of side $B C$, $B, C, H, O$ four points cannot be concyclic, so, $\angle A$ cannot be $90^{\circ}$;

When $\angle A<90^{\circ}$, assume $\angle B=90^{\circ}$, at this time, point $B$ coincides with $H$, thus, $B, C, H, O$ four points are always concyclic, therefore, $\angle A$ can be any angle satisfying $0^{\circ}<\angle A<90^{\circ}$.

In summary, all possible values of $\angle A$ are all acute angles and $120^{\circ}$.