Given a cyclic quadrilateral
As shown:
The measure of the arc FG = 97
The measure of the arc GH = 117
The measure of the arc EHG = 164
The measure of the arc is two times the measure of the inscribed angle opposite to the arc.
So, the measure of the angle E = 1/2 the measure of the arc FGH =
[tex]\frac{1}{2}(\text{arc FG + arc GH ) =}\frac{1}{2}(97+117)=\frac{1}{2}\cdot214=107\degree[/tex]The measure of the angle F = 1/2 the measure of the arc EHG =
[tex]\frac{1}{2}\cdot164=82\degree[/tex]For the cyclic quadrilateral, every two opposite angles are supplementary.
So,
[tex]\begin{gathered} m\angle E+m\angle G=180 \\ m\angle G=180-m\angle E=180-107=73\degree \end{gathered}[/tex]And:
[tex]\begin{gathered} m\angle F+m\angle H=180 \\ m\angle H=180-m\angle F=180-82=98\degree \end{gathered}[/tex]So, the answer will be:
[tex]\begin{gathered} \text{Blank}1\colon107\degree \\ \text{Blank}2\colon82\degree \\ \text{Blank}3\colon73\degree \\ \text{Blank}4\colon98\degree \end{gathered}[/tex]