In cyclic quadrilateral $ABCD$,$m \angle A = 2x + 4$,$m \angle B = y + 3$,$m \angle C = 2y + 10$,and $m \angle D = 4x - 5$. Find the measures of all four angles.

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$m \angle A + m \angle C = 180^{\circ}$ and $m \angle B + m \angle D = 180^{\circ}$.
For $A$ and $C$: $(2x + 4) + (2y + 10) = 180 \implies 2x + 2y + 14 = 180 \implies 2x + 2y = 166 \implies x + y = 83$ (Equation $1$).
For $B$ and $D$: $(y + 3) + (4x - 5) = 180 \implies 4x + y - 2 = 180 \implies 4x + y = 182$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(4x + y) - (x + y) = 182 - 83 \implies 3x = 99 \implies x = 33$.
Substituting $x = 33$ into Equation $1$: $33 + y = 83 \implies y = 50$.
Calculating the angles:
$m \angle A = 2(33) + 4 = 66 + 4 = 70^{\circ}$.
$m \angle B = 50 + 3 = 53^{\circ}$.
$m \angle C = 2(50) + 10 = 100 + 10 = 110^{\circ}$.
$m \angle D = 4(33) - 5 = 132 - 5 = 127^{\circ}$.