10. As shown in Figure 2, given a square $O A B C$ with side length 1 in the Cartesian coordinate system, points $A$ and $B$ are in the first quadrant, and $O A$ makes an angle of $30^{\circ}$ with the $x$-axis. Therefore, the coordinates of point $B$ are $\qquad$
10. $\left(-\frac{1+\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$.

Draw $A D \perp x$-axis at $D$, $B E \perp e$-axis at $E$, $A F \perp B E \Psi$
$F$ is
$$
\begin{array}{l}
F B=\text { (A) } F A=E D=I A \\
\because O A=1 . \angle I O A=30 \\
\therefore F A=E D=\frac{1}{2} .
\end{array}
$$

Also, $O D=\sqrt{O A^{2}-D A^{2}}=\frac{\sqrt{3}}{2}$. Therefore,
$$
\begin{array}{l}
E B=E F+F B=D A+O D=\frac{1+\sqrt{3}}{2}, \\
O E \leftrightharpoons O D \cdots E D=O D-D A=\frac{-1+\sqrt{3}}{2} .
\end{array}
$$