In parallelogram EFGH if GJ=15 find JE.
F
E
H

On line EF,
[tex]Sum\text{ of angles on a straight line =180}^0[/tex]Therefore,
We will have that
[tex]\begin{gathered} x^0+119^0=180^0 \\ x^0=180^0-119^0 \\ x^0=61^0 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} x=\angle H(\text{Alternate angles are equal)} \\ \text{Hence,} \\ \angle H=61^0 \end{gathered}[/tex][tex]\begin{gathered} y=\angle H(corresponding\text{ angles)} \\ \text{therefore,} \\ y=61^0 \end{gathered}[/tex]The relationship between y and angle G IS
[tex]\begin{gathered} y+\angle G=180^0(Angles\text{ on a straight line)} \\ \angle G+61^0=180^0 \\ \angle G=180^0-61^0 \\ \angle G=119^0 \end{gathered}[/tex]In conclusion,
[tex]\begin{gathered} y=\angle F((\text{Alternate angles are equal)} \\ \text{therefore,} \\ \angle F=61^0 \end{gathered}[/tex]Hence,
Angle G=119°
Angle F=61°
Angle H=61°