15. $\triangle A B C$ is a right-angled triangle with $\angle A B C=90^{\circ}$. A circle $C_{1}$ is drawn with $A B$ as diameter, and another circle $C_{2}$ is drawn with $B C$ as diameter. The circles $C_{1}$. and $C_{2}$ meet at the points $B$ and $P$. If $A B=5 \mathrm{~cm}, B C=12 \mathrm{~cm}$ and $B P=x \mathrm{~cm}$, find the value of $\frac{2400}{x}$.
15. Ans: 520
Drop the perpendicular from $B$ to $A C$ meeting $A C$ at $Q$. Then $\angle A Q B=90^{\circ}$, and thus $Q$ lies on $C_{1}$. Similarly, $Q$ lies on $C_{2}$. Thus, $Q=P$. Now, $\triangle A B P \sim \triangle A C B$. Thus
$$
\frac{B P}{A B}=\frac{B C}{A C} \Longrightarrow \frac{x}{5}=\frac{12}{\sqrt{12^{2}+5^{2}}} \Longrightarrow x=\frac{60}{13} \text {. }
$$

Therefore,
$$
\frac{2400}{x}=2400 \times \frac{13}{60}=520 .
$$