Answer
by Theorem one diagonal of a kite bisects two angles of the kite which is $\angle A and \angle C $ therefore $\triangle ABC = \frac{1}{2} \angle A + \frac{1}{2} \angle C + \angle B= 180^{\circ}$
By substitution
$ (\angle B - 30) + (\angle B -50) + \angle B = 180 $
$So \angle B = 110$
Work Step by Step
In the kite $ABCD$, the measures of $\angle B$ and $\angle D$ are equal.
Also, the sum of the four interior angles in a kite is $360^{\circ}$
We can find $m\angle B$:
$m\angle A + m\angle B + m\angle C + m\angle D = 360^{\circ}$
$(m\angle B - 50) + m\angle B + (m\angle B-30) + m\angle B = 360^{\circ}$
$4~m\angle B - 80 = 360^{\circ}$
$4~m\angle B = 440^{\circ}$
$m\angle B = \frac{440^{\circ}}{4}$
$m\angle B = 110^{\circ}$
