ABCD is cyclic quadrilateral. Sides AB and DC, when produced, meet at E, and sides BC and AD, when produced, meet at F. If $\angle$BFA = $60^\circ$ and $\angle$AED = $30^\circ$, then the measure of $\angle$ABC is:

75°

In triangle ABF

\(\alpha \) + \(\theta \) + 60 = 180

= \(\alpha \) + \(\theta \) = 120  ..(1)

In triangle ADE

\(\alpha \) + \(\pi \) - \(\theta \) + 30 = 180

= \(\alpha \) - \(\theta \) + 30 = 0   ..(2)

From eq(1) and (2), we have

\(\alpha \) = 45 and \(\theta \) = 75

Therefore, \(\angle\)ABC = \({75}^\circ\).