The correct option is B 140∘
Intheabovefigure∠BAO=30and∠BCO=40......(1)OA,OBandOCarediameterofcircleHenceOA=OB=OCTheanglesoppositetotheequalsidesareequal⇒∠BAO=∠ABOSimilarly∠BCO=∠CBO⇒∠ABO=30and∠CBO=40[From(1)]Now∠ABC=∠ABO+∠CBO⇒∠ABC=30+40=70.........(2)∠AOC=2∠ABC=2×70=140[From(2)]HenceoptionBisrightanswer