jmc

The sum of the angle measures in a pentagon is $180(5-2)=540$ degrees, so we must have $$\angle M+\angle A+\angle T+\angle H+\angle S=540^{\circ }.$$ Since $\angle A$ and $\angle S$ are supplementary, we have $\angle A+\angle S=180^{\circ }$. Combining this with $\angle H=\angle M=\angle T$, we have $$\begin{array}{rl}\angle M+\angle A+\angle T+\angle H+\angle S& =(\angle M+\angle T+\angle H)\\ & +(\angle A+\angle S)\\ & =3\angle H+180^{\circ },\end{array}$$ so $3\angle H+180^{\circ }=540^{\circ }$. Therefore, $3\angle H=360^{\circ }$ and $\angle H=\boxed{120^{\circ }}$.