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:
Answer & explanation
Correct answer: option 2
\(\angle\)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 \({67}^\circ\).