47. Inside an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $P$ is inside triangle $A Q B$, $P Q = Q C$, and $\angle P A Q = \angle P B Q = 30$. Find $\angle A Q B$.
47. We construct regular triangles $A P X$ and $B P Y$ on the sides $A P$ and $P B$ of triangle $A P B$. Triangle $A X C$ is obtained by rotating triangle $A P B$ by $60^{\circ}$ around point $A$, so these triangles are equal. Therefore, $X C = P B = P Y$. Similarly, $C Y = A P = X P$. Consequently, quadrilateral $P X C Y$ is a parallelogram. Further, since $\angle P B Q = 30^{\circ}$, $B Q$ is the angle bisector of the regular triangle $B P Y$. Therefore, $B Q$ is the perpendicular bisector of $P Y$, and thus $Q P = Q Y$. Similarly, $Q P = Q X$. Since by the condition $Q P = Q C$, we get that point $Q$ is equidistant from all vertices of parallelogram $P X C Y$. Such a point can only be the intersection of its diagonals. But then the diagonals of parallelogram $P X C Y$ are equal, so $P X C Y$ is a rectangle. We have already proven that segments $B Q$ and $A Q$ are perpendicular to the sides of this rectangle. Thus, $\angle A Q B = 90^{\circ}$.