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:


*

*Obviously, the red angles are equal, and the orange angles are equal.  This gives $XY \parallel TZ$.

*$YZ=2PZ$. From this $XY=3TZ$ then $O'Z=3OY$.  Let $O'Z=a=O'S$ so $SZ=\sqrt{2} a$, and also $O'O=2a$

*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.

Triangle O'SO fits the description of a 30-60-90 special angled triangle. Therefore, $\angle O'OS = 30^0$
Then, $\angle PYX = 15^0$ [angles at center = 2 times angles at circumference]
