Answer: A
In triangle BCD: \(\angle\)DBC = \(\angle\)DAC = 70° [lying on the same arc CD] \(\therefore \angle\)BCD = 180° − [\(\angle\)DBC + \(\angle\)BDC] = 180° − [70° + 50°] = 180° − 120° = 60° \(\therefore \angle\)BAD = 180° − 60° = 120° Again, in triangle ABD, AB = AD \(\therefore \angle\)ABD = \(\frac{180^\circ - 120^\circ}{2} = 30^\circ\) \(\therefore \angle\)ACD = \(\angle\)ABD = 30° [lying on the same arc AD]
In triangle BCD: \(\angle\)DBC = \(\angle\)DAC = 70° [lying on the same arc CD] \(\therefore \angle\)BCD = 180° − [\(\angle\)DBC + \(\angle\)BDC] = 180° − [70° + 50°] = 180° − 120° = 60° \(\therefore \angle\)BAD = 180° − 60° = 120° Again, in triangle ABD, AB = AD \(\therefore \angle\)ABD = \(\frac{180^\circ - 120^\circ}{2} = 30^\circ\) \(\therefore \angle\)ACD = \(\angle\)ABD = 30° [lying on the same arc AD]