In the figure below, segments $A B$ and $C D$ are parallel. If $\angle C I J=10 \beta, \angle A G J=10 \alpha$, $\angle C E J=6 \alpha$ and $\angle J F G=6 \beta$, determine the value of the angle $\angle I J K$.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-02.jpg?height=468&width=811&top_left_y=503&top_left_x=523)
Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-02.jpg?height=466&width=825&top_left_y=1229&top_left_x=521)

Since $\angle C E J$ and $\angle J F A$ are consecutive interior angles, $6 \alpha + 6 \beta = 180^{\circ}$, which means $\alpha + \beta = 30^{\circ}$. From point $J$, consider the segment $M L$ parallel to $A B$. We have

\[
\begin{aligned}
x & = 180^{\circ} - \angle I J G \\
& = 180^{\circ} - \angle I J M - \angle M J G \\
& = 180^{\circ} - \angle J I E - \angle J G F \\
& = 180^{\circ} - (180^{\circ} - 10 \beta) - (180^{\circ} - 10 \alpha) \\
& = 10(\alpha + \beta) - 180^{\circ} \\
& = 120^{\circ}
\end{aligned}
\]