\(\angle ADC = \angle XBC = 82^\circ\) [Since when a side of a cyclic quadrilateral is extended, the exterior angle equals the interior opposite angle] ∴ \(\angle BDC = 82^\circ - 47^\circ = 35^\circ\) Again, since \(\angle BDC\) and \(\angle BAC\) lie on the same arc BC of the circle, ∴ \(\angle BAC = \angle BDC = 35^\circ\) [Answer]