28. Two circles $C_{1}$ and $C_{2}$ of radii $10 \mathrm{~cm}$ and $8 \mathrm{~cm}$ respectively are tangent to each other internally at a point $A . A D$ is the diameter of $C_{1}$ and $P$ and $M$ are points on $C_{1}$ and $C_{2}$ respectively such that $P M$ is tangent to $C_{2}$, as shown in the figure below.
If $P M=\sqrt{20} \mathrm{~cm}$ and $\angle P A D=x^{\circ}$, find the value of $x$.
28. Answer: 60

Let $O$ be the centre of $C_{2}$ and let $P A$ intersect $C_{2}$ at $B$. The homothety centred at $A$ mapping $C_{2}$ to $C_{1}$ has similitude ratio $\frac{8}{10}$. It maps $B$ to $P$. Thus $\frac{A B}{A P}=\frac{8}{10}$. (This can also be seen by connecting $P$ to the centre $O^{\prime}$ of $C_{1}$ so that the triangles $A B O$ and $A P O^{\prime}$ are similar.) The power of $P$ with respect to $C_{2}$ is $P M^{2}=20$. Thus $P B \cdot P A=20$, or equivalently $(P A-A B) P \dot{A}=20$. Together with $\frac{A B}{A P}=\frac{8}{10}$, we obtain $A B=8$ and $A P=10$. Consequently, the triangle $A B O$ is equilateral, and hence $\angle P A D=\angle B A O=60^{\circ}$.