Finding an angle in a figure involving tangent circles

Question

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.

Answer 1

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]

Answer 2

Hint:

Notice that $\angle PYX=\angle PYO= \angle OPY$ and $\angle POX=\angle PYO+\angle OPY$, so $$\angle PYX=\frac{1}{2}\angle POX$$ On the other hand, $$\tan \angle POX=\frac{|O'S|}{|SO|}=\frac{a}{a\sqrt{3}}=\frac{1}{\sqrt{3}}\qquad\implies\qquad \angle POX=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$$