# Finding an angle in a figure involving tangent circles

The circle $A$ touches the circle $B$ internally at $P$. The centre $O$ of $B$ is outside $A$. Let $XY$ be a diameter of $B$ which is also tangent to $A$. Assume $PY > PX$. Let $PY$ intersect $A$ at $Z$. If $Y Z = 2PZ$, what is the magnitude of $\angle PYX$ in degrees?

What I have tried:

1. Obviously, the red angles are equal, and the orange angles are equal. This gives $XY \parallel TZ$.
2. $YZ=2PZ$. From this $XY=3TZ$ then $O'Z=3OY$. Let $O'Z=a=O'S$ so $SZ=\sqrt{2} a$, and also $O'O=2a$
3. Then $SO=\sqrt{3} a$. Now we can use trigonometry to find $\angle PYX$ in triangle $ZSY$. Please verify whether my figure is correct. Your solution to this question is welcomed, especially if it is shorter.

• Why do we have $SO=3O'Z$? From $3TZ=XY$ it follows that $XO=3O'Z$, so we can't possible have $SO=3O'Z$... – math635 Dec 23 '15 at 4:40
• Yah I have edited now – mnulb Dec 23 '15 at 4:44
• It's wrong again. Do check what you write before you ask other people. – user21820 Dec 23 '15 at 4:54
• Your solution looks more or less correct to me, but i have to concerns. 1. How do you find $ZY$ in terms of $a$? 2. How can you use trigonometry to determine$\angle PYX$ – math635 Dec 23 '15 at 5:02