15. $\frac{\sqrt{21}}{3}$
15. Note that $\triangle A B C$ and $\triangle A C D$ are equilateral. Rotate $\triangle A C P$ clockwise through $60^{\circ}$ about $C$ to $\triangle D C Q$. Note that $\triangle P C Q$ is equilateral and hence $A P Q$ is a straight line. Also, we have $\triangle A Q C \cong \triangle B P C$. Since $\angle A P C+\angle A B C=180^{\circ} \quad, \quad A B C P$ is a cyclic quadrilateral. Hence $\angle D Q P=\angle A P B=\angle A C B=60^{\circ}$.
Let $A P=x$ and $C P=y$. Then
$$
3=B P=A Q=A P+P Q=x+y .
$$

On the other hand, applying cosine law in $\triangle D Q P$ gives $2^{2}=x^{2}+y^{2}-2 x y \cos 60^{\circ}$. These two equations give $x^{2}+y^{2}+2 x y=9$ and $x^{2}+y^{2}-x y=4$ respectively. Thus $x y=\frac{5}{3}$ and hence $x^{2}+y^{2}-2 x y=4-\frac{5}{3}=\frac{7}{3}$. Thus $(x-y)^{2}=\frac{3}{7}$, or $(x-y)=\sqrt{\frac{7}{3}}$, i.e. the difference between the lengths of $A P$ and $C P$ is $\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3}$.
15. $A B C D$ is a rhombus with $\angle B=60^{\circ} . P$ is a point inside $A B C D$ such that $\angle A P C=120^{\circ}$, $B P=3$ and $D P=2$. Find the difference between the lengths of $A P$ and $C P$.
(2 marks)
$A B C D$ 是菱形, 當中 $\angle B=60^{\circ} \circ P$ 是 $A B C D$ 內的一點, 使得 $\angle A P C=120^{\circ} 、 B P=$ $3 、 D P=2$ 。求 $A P$ 與 $C P$ 的長度之差。