For a better understanding of the solution given here please find the attached file which has the diagram of the kite EFGH.
As can be seen from the diagram, [tex] \angle FEH=\angle FGH=102^{\circ} [/tex]
Also, it is given that [tex] \angle EHG=68^{\circ} [/tex]
We are required to find the [tex] \angle EFG [/tex].
To proceed we know that a kite is a quadrilateral and in a quadrilateral the sum of the interior angles is 360 degrees. Thus, we will get:
[tex] m\angle FEH+m\angle EHG+m\angle FGH+m\angle EFG=360^{\circ} [/tex]
Plugging in the given values we get:
[tex] 102^{\circ}+68^{\circ}+102^{\circ}+m\angle EFG=360^{\circ} [/tex]
This will give:
[tex] m\angle EFG=88^{\circ} [/tex]
Therefore, Option B is the correct option.
