Answer: A
In triangle ABC, \(\angle\)ABC = 180° – (\(\angle\)BAC + \(\angle\)BCA) = 180° – (85° + 75°) = 20° \(\therefore\) Central angle \(\angle\)AOC = 2 × \(\angle\)ABC = 2 × 20° = 40° Now, in triangle AOC, AO = OC \(\therefore \angle\)OAC = \(\angle\)OCA = \(\frac{1}{2}\)(180° – 40°) = 70° \(\therefore \angle\)OAC = 70° (Answer)
In triangle ABC, \(\angle\)ABC = 180° – (\(\angle\)BAC + \(\angle\)BCA) = 180° – (85° + 75°) = 20° \(\therefore\) Central angle \(\angle\)AOC = 2 × \(\angle\)ABC = 2 × 20° = 40° Now, in triangle AOC, AO = OC \(\therefore \angle\)OAC = \(\angle\)OCA = \(\frac{1}{2}\)(180° – 40°) = 70° \(\therefore \angle\)OAC = 70° (Answer)