Help with this right angle triangle

[specialvalue]

The triangle: <link removed due to objectionable content>

I know angle ABC is 60° (there's a previous exercise).
I'm asked to find sin(x).
So, I used cos(60°) to find side AB is 9 and then side CA is 9*root(3) by Pythagoras. Then I tried the cosine rule to find side AN and got 3*root(7). Then since ACB is 30° side AM is 3*root(13) and cos(x) should be 8/(9*root(91)). But now I'm not sure how to proceed, I tried making a right angle triangle since I have cos(x) but I got a huge mess.
Any help will be appreciated.

 

[stapel]

For other readers: Here's a textual version of the picture:

C*
 | \ 6
 |   \.P
 |   / \ 6 
 |  /    \.Q
 | /   .'  \
 |/ .'       \
A*------------* B

The angle at A is 90 degrees; the angle at B is 60 degrees. The lengths of segments CP, PQ, and QB are equal.