In a cyclic quadrilateral ABCD, if $\angle A=\left(2x+4\right)°,\angle B=\left(y+3\right)°,\angle C=\left(2y+10\right)°$  and $\angle D=\left(4x-5\right)°$ , then $y=$

In a cyclic quadrilateral the sum of opposite angles in equal to $180°$

$\angle B+\angle D=180°$

$y+3+4x-5=180°$

$4x+y=182°\to \left(1\right)$

$\angle A+\angle C=180°$

$2x+4+2y+10=180°$

$2x+2y=180°-14$

$2x+2y=166$

$x+y=83\to \left(2\right)$

Solve (1) and (2)

$x=33$

The value of $\text{'}x\text{'}$ is substituting in equation (2)

$33+y=83$

$y=83-33$

$y=50°$