Example 1 See Figure 15-1, after joining a set of triangular rulers, let triangular ruler $BCD$ stand up along $BC$, making the planes of the two triangular rulers perpendicular to each other, find the angle formed by the skew lines $AD$ and $BC$.
Given $\angle B A C=\angle B C D=90^{\circ}, \angle C B A=45^{\circ}$ and $\angle B D C=60^{\circ}$
Let $O$ be the midpoint of $B C$ as the origin, and establish a spatial rectangular coordinate system, with $O C=1$
Then point $C(0,1,0)$, point $A(1,0,0)$, point $D\left(0,1, \frac{2}{\sqrt{3}}\right)$
So $\overrightarrow{A D}=\left(0,1, \frac{2}{\sqrt{3}}\right)-(1,0,0)=\left(-1,1, \frac{2}{\sqrt{3}}\right)$
$$
\overrightarrow{A D} \cdot \overrightarrow{O C}=1
$$

So $\operatorname{cos} \langle \overrightarrow{A D}, \overrightarrow{O C}\rangle=\frac{\overrightarrow{A D} \cdot \overrightarrow{O C}}{|\overrightarrow{A D}| \cdot|\overrightarrow{O C}|}=\frac{1}{\sqrt{(-1)^{2}+1^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}}=\sqrt{\frac{3}{10}}$ Therefore, the angle between the skew lines $A D$ and $B C$ is $\arccos \sqrt{\frac{3}{10}}$