Geometry problem with circles and arcs In the diagram, $\angle U = 30^\circ$, arc $XY$ is $170^\circ$, and arc $VW$ is $110^\circ$. Find arc $WY$, in degrees.

I got $55^\circ$ but apparently it's wrong.
 A: Draw the figure such that $XY$ is horizontal. Then draw provisional $V'W'$ forming an angle of $110^\circ$ at the center $O$ of the circle, so that $V'W'$ is horizontal as well. Now  rotate $V'W'$ clockwise into the definitive position $VW$ so that at $U$ you obtain the angle $30^\circ$. Looking at all angles and half-angles in your figure then shows you that the angle you desire is $70^\circ$.
A: Use this fact about two secants that meet at a point outside the circle:
$\newcommand{arc}{\mathop{\mathrm{arc}}}$
$$ \frac12(\arc WY - \arc XV) = \angle U = 30^\circ. $$
Also use the fact that the circle is divided into four arcs:
$$ \arc WY + \arc XV = 360^\circ - \arc XY - \arc VW
 = 360^\circ - 170 ^\circ - 110^\circ = 80^\circ. $$
Now you have two equations in just two unknowns, $\arc WY$ and $\arc XV.$
Solve for $\arc WY.$
A: 
Construct $YV$ as the diameter of the circle. Then
\begin{align}
\angle XOY&=170^\circ
,\\
\angle XWY&=85^\circ
,\\
\angle XYO=\angle OYX=\angle VWX
&=5^\circ
,\\
\angle VOX&=10^\circ
,\\
\angle YOV&=70^\circ
,\\
\angle VYU&=60^\circ
,\\
\angle UWY&=90^\circ
,\\
\angle YUW&=30^\circ
.
\end{align}
Edit
Alternatively,

\begin{align}
\triangle UWY:\quad
\angle YUW&+\angle UWY+\angle WYU
=180^\circ
,\\
\angle UWY&=
\angle UWO+\angle OWY
=125^\circ-\tfrac12\phi
,\\
\angle WYU&=
\angle WUO+\angle OYU
=95^\circ-\tfrac12\phi
,\\
30^\circ+
125^\circ&-\tfrac12\phi
+95^\circ-\tfrac12\phi
=180^\circ
,\\
\phi&=70^\circ
.
\end{align}
