Рассмотрим треугольник BFC. \(\angle BFC = 180^\circ - 70^\circ = 110^\circ\). \(\angle FBC + \angle FCB + \angle BFC = 180^\circ\). \(30^\circ + 20^\circ + 110^\circ = 160^\circ\). Значит, \(\angle FBC = 30^\circ\), \(\angle FCB = 20^\circ\), \(\angle BFC = 70^\circ\).
Рассмотрим треугольник ABC. \(\angle ABC = 30^\circ * 2 = 60^\circ\). \(\angle ACB = 20^\circ * 2 = 40^\circ\). \(\angle BAC + \angle ABC + \angle ACB = 180^\circ\). \(\angle BAC + 60^\circ + 40^\circ = 180^\circ\), \(\angle BAC = 80^\circ\).
**Ответ: \(\angle A = 80^\circ\)**