1. **Stating the problem:** We need to find the measure of angle $m\angle G$ in quadrilateral $EFGH$. 2. **Given information:** - $m\angle F = 150^\circ$. - $EF = FG$ and $EH = HG$ (two pairs of equal sides). - $\angle H$ is a right angle, so $m\angle H = 90^\circ$. 3. **Important rules:** - The sum of interior angles in any quadrilateral is $360^\circ$. - Since $EF = FG$, triangle $EFG$ is isosceles with $\angle E = \angle G$. - Since $EH = HG$, triangle $EHG$ is isosceles with $\angle E = \angle G$ in that triangle as well. 4. **Find $m\angle E$ and $m\angle G$ in triangle $EFG$:** - In triangle $EFG$, $m\angle F = 150^\circ$. - Sum of angles in triangle $EFG$ is $180^\circ$. - Let $m\angle E = m\angle G = x$. $$x + x + 150^\circ = 180^\circ$$ $$2x = 180^\circ - 150^\circ$$ $$2x = 30^\circ$$ $$x = \frac{30^\circ}{2} = 15^\circ$$ So, $m\angle E = m\angle G = 15^\circ$. 5. **Check angles in quadrilateral $EFGH$:** - $m\angle F = 150^\circ$ - $m\angle H = 90^\circ$ - $m\angle E = 15^\circ$ - $m\angle G = ?$ Sum of angles in quadrilateral: $$m\angle E + m\angle F + m\angle G + m\angle H = 360^\circ$$ Substitute known values: $$15^\circ + 150^\circ + m\angle G + 90^\circ = 360^\circ$$ Simplify: $$255^\circ + m\angle G = 360^\circ$$ Solve for $m\angle G$: $$m\angle G = 360^\circ - 255^\circ = 105^\circ$$ **Final answer:** $$m\angle G = 105^\circ$$