24. Answer. 30
Let $P O_{1}=r_{1}$ and $P O_{2}=r_{2}$. First note that $O_{1} O_{2}$ intersects $P Q$ at the midpoint $H$ (not shown in the figure) of $P Q$ perpendicularly. Next observe that $\angle A P Q=\angle P B Q=$ $\angle P O_{2} O_{1}$, and $\angle B P Q=\angle P A Q=\angle P O_{1} O_{2}$. Therefore $\angle A P B=\angle A P Q+\angle B P Q=$ $\angle P O_{2} O_{1}+\angle P O_{1} O_{2}$.

Let $\angle P O_{2} O_{1}=\alpha$ and $\angle P O_{1} O_{2}=\beta$. Then $\sin \alpha=\frac{P Q}{2 r_{2}}, \cos \alpha=\frac{O_{2} H}{r_{2}}$ and $\sin \beta=$ $\frac{P Q}{2 r_{1}}, \cos \beta=\frac{O_{1} H}{P_{2}}$. Thus $\sin \angle A P B=\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta=\frac{P Q}{2 r_{2}} \cdot \frac{O_{1} H}{r_{1}}+$ $\frac{O_{1} H}{r_{2}} \cdot \frac{P Q}{2 r_{1}}=\frac{P Q \cdot\left(O_{1} H+O_{2} H\right)}{2 r_{1} r_{2}}=\frac{P Q \cdot O_{1} O_{2}}{2 r_{1} r_{2}}=\frac{1}{2}$. Since $\angle A P B$ is acute, it is equal to $30^{\circ}$.
24. The circle $\gamma_{1}$ centred at $O_{1}$ intersects the circle $\gamma_{2}$ centred at $O_{2}$ at two points $P$ and $Q$. The tangent to $\gamma_{2}$ at $P$ intersects $\gamma_{1}$ at the point $A$ and the tangent to $\gamma_{1}$ at $P$ intersects $\gamma_{2}$ at the point $B$ where $A$ and $B$ are distinct from $P$. Suppose $P Q \cdot O_{1} O_{2}=P O_{1} \cdot P O_{2}$ and $\angle A P B$ is acute. Determine the size of $\angle A P B$ in degrees.