First, notice that $\angle{JIC}$ and $\angle{EJA}$ are corresponding angles. Therefore, their measures are equal. \[ m\angle{EJA} = m\angle{JIC} = 2x + 4^\circ \]

Also, the angles $\angle{EJA}$ and $\angle{EJB}$ are supplementary. So, we have:

\begin{align*} m\angle{EJA} + m\angle{EJB} &= 180^\circ \\[5pt] (3x+11^\circ) + (2x+4^\circ) &= 180^\circ \\[5pt] 5x+15^\circ &= 180^\circ \\[5pt] 5x &= 165^\circ \\[5pt] x &= 33^\circ \end{align*}

First, notice that $\angle{EJA}$ and $\angle{EIC}$ are corresponding angles. Therefore, their measures are equal. \[ m\angle{EJA} = m\angle{EIC} = 4x + 12^\circ \]

Also, the angles $\angle{EID}$ and $\angle{EIC}$ are supplementary. So, we have:

\begin{align*} m\angle{EID} + m\angle{EIC} &= 180^\circ \\[5pt] (16x-42^\circ) + (14x+12^\circ) &= 180^\circ \\[5pt] 30x-30^\circ &= 180^\circ \\[5pt] 30x &= 210^\circ \\[5pt] x &= 7^\circ \end{align*}

We notice that $\angle{EID}$ and $\angle{BJF}$ are consecutive interior angles. Therefore, they are supplementary, and the sum of their measures is $180^\circ.$

So, we have:

\begin{align*} m\angle EID + m\angle BJF &= 180^\circ \\[5pt] 110^\circ + y &= 180^\circ \\[5pt] y &= 70^{\circ} \end{align*}

We notice that $\angle{JID}$ and $\angle{IJB}$ are consecutive interior angles. Therefore, they are supplementary, and the sum of their measures is $180^\circ.$

So, we have:

\begin{align*} m\angle{JID} + \angle{IJB} &= 180^\circ \\[5pt] \left(x+1^\circ\right)+\left(4x-6^\circ\right) &= 180^\circ \\[5pt] 5x-5^\circ &= 180^\circ \\[5pt] 5x &= 185^\circ \\[5pt] x &= 37^\circ \end{align*}

We notice that $\angle{FID}$ and $\angle{EJB}$ are consecutive exterior angles. Therefore, they are supplementary, and the sum of their measures is $180^\circ.$

So, we have:

\begin{align*} m\angle{FID} + \angle{EJB} &= 180^\circ \\[5pt] 45^\circ + (x-5^\circ) &= 180^\circ \\[5pt] x+40^\circ &= 180^\circ \\[5pt] x &= 140^{\circ} \end{align*}

We notice that $\angle{FID}$ and $\angle{EJB}$ are consecutive exterior angles. Therefore, they are supplementary, and the sum of their measures is $180^\circ.$

So, we have:

\begin{align*} m\angle{FID} + \angle{EJB} &= 180^\circ \\[5pt] \left(x+30^\circ\right) + 2x &= 180^\circ \\[5pt] 3x+30^\circ &= 180^\circ \\[5pt] 3x &= 150^\circ \\[5pt] x &= 50^\circ \end{align*}

Consider $\angle 2 $ and $\angle 3$ in the diagram below.

Note that $\angle 2$ and the angle whose measure is $40^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 2 = 40^\circ. \]

Similarly, note that $\angle 3$ and the angle whose measure is $129^\circ$ are consecutive exterior angles formed by the same transversal. So, they are supplementary, and the sum of their measures is $180^\circ.$ Therefore:

\begin{align*} m\angle 3 + 129^\circ &= 180^\circ \\[3pt] m\angle 3 &= 51^\circ \end{align*}

The angles $\angle 1,$ $\angle 2,$ and $\angle 3$ combine to give a straight angle. So, we get:

\begin{align} m\angle 1 +m\angle 2+ m\angle 3 &= 180^\circ\\[5pt] m\angle 1 + 40^\circ + 51^\circ &= 180^\circ\\[5pt] m\angle 1 + 91^\circ &= 180^\circ\\[5pt] m\angle 1 &= 89^\circ \end{align}

Consider $\angle 4$ and $\angle 5$ in the diagram below.

First, notice that $\angle{2}$ and $\angle{4}$ are consecutive interior angles formed by the same transversal. Therefore, \begin{align*} m \angle{4} &= 180^{\circ} - m\angle{2} \\[5pt] &= 180^{\circ}-\left(x+19^{\circ}\right) \\[5pt] &= 161^{\circ}-x. \end{align*}

Now, notice that $\angle{5}$ and $\angle{1}$ are alternate interior angles formed by another transversal. Therefore, \begin{align*} m\angle{5} &=m \angle{1} \\[5pt] &= 4x-10^{\circ}. \end{align*}

Finally, angles $\angle{3},$ $\angle{4},$ and $\angle{5}$ form a straight angle. So, we obtain:

\begin{align*} m\angle{3} + m\angle{4} + m\angle{5} &= 180^{\circ} \\[5pt] \left(2x + 9^{\circ} \right) + \left(161^{\circ}-x\right) + \left(4x-10^{\circ}\right) &= 180^{\circ} \\[5pt] 5x + 160^{\circ} &= 180^{\circ} \\[5pt] 5x &= 20^{\circ} \\[5pt] x &= 4^{\circ} \end{align*}