Finding a triangle ratio. 
In the triangle ABC, the point P is found on the side AB. AC = 6 cm, AP = 4 cm, AB = 9 cm. Calculate BC:CP.

For some reason, I cannot get this even though I tried for half an hour.
I got that, $BC/CP < 17/10 = 1.7$ by the triangle inequality. $AP/PB = 4/5$
But that does not help one but, I'm very stuck!

Let $\angle{BAC}=\theta$. Then, by the law of cosines, 
$$\begin{align}BC:CP&=\sqrt{9^2+6^2-2\cdot 9\cdot 6\cos\theta}:\sqrt{4^2+6^2-2\cdot 4\cdot 6\cos\theta}\\&=\sqrt{9(13-12\cos\theta)}:\sqrt{4(13-12\cos\theta)}\\&=\color{red}{3:2}\end{align}$$
