1. As shown in Figure 1, given $\angle A O B=90^{\circ}, \angle B=30^{\circ}$, $\triangle A^{\prime} O B^{\prime}$ can be considered as
the result of rotating $\triangle A O B$ around point $O$
clockwise by an angle $\alpha$.
If point $A^{\prime}$ lies on
$A B$, then the size of $\alpha$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
- 1. C.

It is known that $\angle A=60^{\circ}$.
Since $O A^{\prime}=O A$, we have $\angle A^{\prime} O A=\angle O A^{\prime} A=\angle A=60^{\circ}$. Therefore, $\alpha=60^{\circ}$.