In a quadrilateral ABCD, the bisectors of $\angle C$ and $\angle D$ meet at point E. If $\angle C E D=57^{\circ}$ and $\angle A=47^{\circ}$, then the measure of $\angle B$ is:

67°

\(\angle\)CED = \({57}^\circ\) and \(\angle\)A = \({47}^\circ\)

The bisectors of \(\angle\)C and \(\angle\)D meet at point E

\(\angle\)A + \(\angle\)B = 2\(\angle\)CED

⇒ \({47}^\circ\) + \(\angle\)B = 2 x \({57}^\circ\)

⇒ \(\angle\)B = \({114}^\circ\) - \({47}^\circ\) = \({67}^\circ\)

Therefore, \(\angle\)B is \({67}^\circ\).