II. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, there is an object $A$ flying at a speed of $0.3 \mathrm{~km} / \mathrm{s}$ parallel to the positive direction of the $x$-axis. When object $A$ is at point $N$, it is measured that $O N=10 \mathrm{~km}$, $\angle x O N=105^{\circ}$. At this moment, a missile $B$ is launched from the origin $O$ at a speed of $0.7 \mathrm{~km} / \mathrm{s}$ along a straight line. To make the missile $B$ hit object $A$, find the launch angle $\angle x O B$ (accurate to $0.001^{\circ}$).
II. Suppose after $t \mathrm{~s}$, missile $B$ precisely hits the target at point $C$ (as shown in Figure 2),
and let $\angle x O B=\alpha$. Then
$$
\begin{array}{l}
N C=0.3 t, \\
O C=0.7 t, \\
\angle N O C=105^{\circ}-\alpha, \\
\angle O N C=75^{\circ} .
\end{array}
$$

In $\triangle O N C$, by the Law of Sines we get
$$
\begin{array}{l}
\frac{N C}{\sin \angle N O C}=\frac{O C}{\sin \angle O N C} \\
\Rightarrow \frac{0.3 t}{\sin \left(105^{\circ}-\alpha\right)}=\frac{0.7 t}{\sin 75^{\circ}} \\
\Rightarrow \sin \left(105^{\circ}-\alpha\right)=\frac{3}{7} \sin 75^{\circ} \\
\Rightarrow 105^{\circ}-\alpha=\arcsin \left(\frac{3}{7} \sin 75^{\circ}\right)
\end{array}
$$

or $105^{\circ}-\alpha=180^{\circ}-\arcsin \left(\frac{3}{7} \sin 75^{\circ}\right)$
$\Rightarrow \alpha=80.546^{\circ}$ or $\alpha=-50.546^{\circ}$ (discard) Therefore, the required launch angle
$$
\angle x O B=80.546^{\circ} \text {. }
$$