1. **Stating the problem:** We need to find the measure of angle $\angle G$ in quadrilateral $EFGH$ where $\angle F = 150^\circ$, $EF \cong FG$, and $\angle H$ is a right angle ($90^\circ$). 2. **Important properties:** Since $EF \cong FG$, triangle $EFG$ is isosceles with $\angle E = \angle G$. 3. **Sum of angles in triangle $EFG$:** The sum of interior angles in any triangle is $180^\circ$. So, $$\angle E + \angle F + \angle G = 180^\circ$$ 4. **Using isosceles property:** Since $\angle E = \angle G$, let $\angle G = x$. Then, $$x + 150^\circ + x = 180^\circ$$ 5. **Simplify and solve for $x$:** $$2x + 150^\circ = 180^\circ$$ $$2x = 180^\circ - 150^\circ$$ $$2x = 30^\circ$$ $$x = \frac{30^\circ}{2}$$ $$x = 15^\circ$$ 6. **Conclusion:** The measure of $\angle G$ is $15^\circ$. **Final answer:** $$m\angle G = 15^\circ$$