Since the angles $\angle{FIC}$ and $\angle{IJA}$ are corresponding angles, they are congruent and therefore have the same measure:

\begin{align*} m\angle IJA &= m\angle FIC\\[5pt] y &= 105^\circ \end{align*}

Since the angles $\angle{FID}$ and $\angle{IJB}$ are corresponding angles, they are congruent and therefore have the same measure:

\begin{align*} m\angle{FID} &=m\angle{IJB} \\[5pt] 60^\circ &= 8x-4^\circ \end{align*}

Now, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} 60^\circ &= 8x-4^\circ \\[5pt] 64^\circ &= 8x \\[5pt] \dfrac{64^\circ}{8} &= \dfrac{8x}{8} \\[5pt] 8^\circ &= \dfrac{\cancel{8}x}{\cancel{8}} \\[5pt] 8^\circ &= x \\[5pt] x &= 8^{\circ} \end{align*}

Since the angles $\angle{JID}$ and $\angle{EJB}$ are corresponding angles, they are congruent and therefore have the same measure:

\begin{align*} m\angle{JID} &= m\angle{EJB} \\[5pt] 14x-1^\circ &= 12x+17^\circ \end{align*}

Now, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} 14x-1^\circ &= 12x+17^\circ \\[5pt] 14x-12x &= 17^\circ +1^\circ\\[5pt] 2x &= 18^\circ \\[5pt] \dfrac{2x}{2} &= \dfrac{18^\circ}{2} \\[5pt] \dfrac{\cancel{2}x}{\cancel{2}} &= 9^{\circ} \\[5pt] x &= 9^{\circ} \end{align*}

Since the angles $\angle{AJF}$ and $\angle{CIF}$ are corresponding angles, they are congruent and therefore have the same measure:

\begin{align*} m\angle{AJF} &= m\angle{CIF} \\[5pt] 6x+ 10^\circ &= x + 90^\circ \end{align*}

Now, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} 6x + 10^\circ &= x + 90^\circ \\[5pt] 6x - x &= 90^\circ - 10^\circ\\[5pt] 5x &= 80^\circ \\[5pt] \dfrac{5x}{5} &= \dfrac{80^\circ}{5} \\[5pt] \dfrac{\cancel{5}x}{\cancel{5}} &= 16^{\circ} \\[5pt] x &= 16^{\circ} \end{align*}

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

Note that $\angle 2$ and the angle whose measure is $63^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 2 = 63^\circ. \]

Similarly, note that $\angle 3$ and the angle whose measure is $58^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 3 = 58^\circ. \]

Let's add these measures to our diagram:

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

\begin{align} m\angle w +m\angle 2+ m\angle 3&= 180^\circ \\[3pt] m\angle{w} + 63^\circ +58^\circ &= 180^\circ \\[3pt] m\angle{w} + 121^\circ &= 180^\circ \\[3pt] m\angle{w} &= 59^\circ \end{align}

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

Note that $\angle 2$ and the angle whose measure is $103^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 2 = 103^\circ. \]

Similarly, note that $\angle 3$ and the angle whose measure is $37^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 3 = 37^\circ. \]

Let's add these measures to our diagram:

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

\begin{align} m\angle 1 +m\angle 2+ m\angle 3 &= 180^\circ \\[5pt] m\angle{1} + 103^\circ +37^\circ &= 180^\circ \\[5pt] m\angle{1} + 140^\circ &= 180^\circ \\[5pt] m\angle{1} &= 40^\circ \end{align}

Consider $\angle 1$ and $\angle 2$ in the diagram below.

Note that $\angle 1$ and the angle whose measure is $14x$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 1 = 14x. \]

Similarly, note that $\angle 2$ and the angle whose measure is $3x+22^\circ$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 2 = 3x+22^\circ. \]

Let's add these measures to our diagram.

The angles $\angle 1$, $\angle 2,$ and the third angle whose measure is $73^\circ,$ combine to give a straight angle. So, we get:

\begin{align*} \require{cancel} m \angle 1 + m\angle 2 + 73^{\circ} &= 180^{\circ}\\[5pt] 14x+(3x+22^{\circ})+73^{\circ} &= 180^{\circ}\\[5pt] 17x+95^{\circ} &= 180^{\circ}\\[5pt] 17x &=85^{\circ}\\[5pt] \dfrac{17x}{17} &= \dfrac{85^{\circ}}{17}\\[5pt] \dfrac{\cancel{17}x}{\cancel{17}} &= 5^{\circ}\\[5pt] x &=5^{\circ} \end{align*}

Consider $\angle 1$ and $\angle 2$ in the diagram below.

Note that $\angle 1$ and the angle whose measure is $3x$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 1 = 3x. \]

Similarly, note that $\angle 2$ and the angle whose measure is $4x+5^{\circ}$ are corresponding angles formed by the same transversal. Therefore,

\[ m\angle 2 = 4x+5^{\circ}. \]

Let's add these measures to our diagram.

The angles $\angle 1$, $\angle 2,$ and the third angle whose measure is $70^\circ$, combine to give a straight angle. So, we get:

\begin{align*} \require{cancel} m \angle 1 + m\angle 2 + 70^\circ &=180^{\circ}\\[5pt] 3x + (4x+5^{\circ}) + 70^\circ &=180^{\circ}\\[5pt] 7x + 75^{\circ} &=180^{\circ}\\[5pt] 7x &=105^{\circ}\\[5pt] \dfrac{7x}{7} &=\dfrac{105^{\circ}}{7}\\[5pt] \dfrac{\cancel{7}x}{\cancel{7}} &= 15^{\circ}\\[5pt] x &= 15^{\circ} \end{align*}