9. In the following diagram, $A B C D$ is a square with $P A=a, P B=2 a$ and $P C=3 a$. Find $\angle A P B$.
(A) $120^{\circ}$
(B) $130^{\circ}$
(C) $135^{\circ}$
(D) $140^{\circ}$
(E) $145^{0}$
9. (C)
Let $A B=x$ and $\angle P A B=\alpha$. By using Cosine Rule and Pythagoras' Theorem we have
$$
\begin{array}{l}
x^{2}+a^{2}-2 a x \cos \alpha=(2 a)^{2} \\
(x-a \cos \alpha)^{2}+(x-a \sin \alpha)^{2}=(3 a)^{2} .
\end{array}
$$

We obtain from the above two equations that
$$
\cos \alpha=\frac{x^{2}-3 a^{2}}{2 a x} \text { and } \sin \alpha=\frac{x^{2}-5 a^{2}}{2 a x} .
$$

Using $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we have
$$
\left(x^{2}-3 a^{2}\right)^{2}+\left(x^{2}-5 a^{2}\right)^{2}=4 a^{2} x^{2}
$$

Therefore $x^{4}-10 a^{2} x^{2}+17 a^{4}=0$.
Solving, we have $x^{2}=(5 \pm 2 \sqrt{2}) a^{2}$, but since $x>a$, we have $x^{2}=(5+2 \sqrt{2}) a^{2}$.
Hence $\cos \angle A P B=-\frac{\sqrt{2}}{2}$, so the required angle is $135^{\circ}$