2.2. $A B C$ is an equilateral triangle. On one side of the plane $A B C$, perpendicular segments $A A^{\prime}=A B$ and $B B^{\{\prime}}=A B / 2$ are laid off. Find the angle between the planes $A B C$ and $A^{\prime} B^{\prime} C$.
2.2. Answer: $45^{\circ}$.

Note that the line $A^{\prime} B^{\prime}$ intersects the plane $A B C$ at a point $D$ such that $\overrightarrow{B D}=\overrightarrow{A B}$. It is clear that the planes $A B C$ and $A^{\prime} B^{\prime} C$ intersect along the line $C D$. It is also easy to see that $C D \perp A C$. Therefore, $C D \perp A^{\prime} C$. By the definition of the angle between planes, it is equal to the angle between $A^{\prime} C$ and $A C$. This is the hypotenuse and one leg of a right isosceles triangle.