The angles of a cyclic quadrilateral $ABCD$ are $\angle A = (6x + 10)^{\circ}$,$\angle B = (5x)^{\circ}$,$\angle C = (x + y)^{\circ}$,and $\angle D = (3y - 10)^{\circ}$. Find $x$ and $y$,and hence the values of the four angles.

(A) We know that,by the property of a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
$\angle A + \angle C = (6x + 10)^{\circ} + (x + y)^{\circ} = 180^{\circ}$
$\Rightarrow 7x + y = 170 \quad \dots(i)$
Similarly,$\angle B + \angle D = (5x)^{\circ} + (3y - 10)^{\circ} = 180^{\circ}$
$\Rightarrow 5x + 3y = 190 \quad \dots(ii)$
Multiplying equation $(i)$ by $3$,we get $21x + 3y = 510 \quad \dots(iii)$
Subtracting equation $(ii)$ from $(iii)$:
$(21x + 3y) - (5x + 3y) = 510 - 190$
$16x = 320 \Rightarrow x = 20$
Substituting $x = 20$ in equation $(i)$:
$7(20) + y = 170 \Rightarrow 140 + y = 170 \Rightarrow y = 30$
Now,calculating the angles:
$\angle A = 6(20) + 10 = 130^{\circ}$
$\angle B = 5(20) = 100^{\circ}$
$\angle C = 20 + 30 = 50^{\circ}$
$\angle D = 3(30) - 10 = 80^{\circ}$
Thus,$x = 20$,$y = 30$,and the angles are $130^{\circ}, 100^{\circ}, 50^{\circ}, 80^{\circ}$.