Answer: A
In triangle ABC, \(\angle\)ABC = 180° − (\(\angle\)BAC + \(\angle\)BCA) = 180° − (85° + 70°) = 25° ∴ Central angle \(\angle\)AOC = 2 × \(\angle\)ABC = 2 × 25° = 50° Now, in triangle AOC, AO = OC ∴ \(\angle\)OAC = \(\angle\)OCA = \(\frac{1}{2}\)(180° − 50°) = 65° ∴ \(\angle\)OAC = 65° (Answer)
In triangle ABC, \(\angle\)ABC = 180° − (\(\angle\)BAC + \(\angle\)BCA) = 180° − (85° + 70°) = 25° ∴ Central angle \(\angle\)AOC = 2 × \(\angle\)ABC = 2 × 25° = 50° Now, in triangle AOC, AO = OC ∴ \(\angle\)OAC = \(\angle\)OCA = \(\frac{1}{2}\)(180° − 50°) = 65° ∴ \(\angle\)OAC = 65° (Answer)