In rectangle $EFGH$, let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$, as shown below. If $\angle EFH = 30^\circ$ and $\angle MHX = 60^\circ,$ then find $\angle XHG,$ in degrees.
... and let $X$ be a point such that $MH = MX$, as shown below
Did you forget to attach a diagram?
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