In the given figure, O is the centre of the circle ∠OAB = 30o and ∠OCB = 40o. Then ∠AOC is
A
70o
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B
20o
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C
60o
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D
120o
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Solution
The correct option is A70o Given−Oisthecentreofacircle.OnitscircumferencethreepointsA,B&Caresuchthat∠OAB=30o&∠OCA=40o.Tofindout−∠AOC=?Solution−WejoinAC.InΔOACwehaveOA=OC(radiiofthesamecircle).∴ΔOACisisosceleswithACasbase.∴∠OAC=∠OCA=x(say).∴∠AOC=180o−(x+x)=180o−2x.........(i).NowthechordACsubtends∠AOCtothecentreOand∠ABCtothecircumferenceatB.∴∠AOC=2∠ABC⟹∠ABC=12∠AOC=12×(180o−2x)=90o−x(fromi).......(ii).AgaininΔABCwehave∠BAC=∠OAB+∠OAC=30o+xand∠BCA=∠OCA+∠OCB=40o+x.∴∠ABC=180o−(∠BAC+∠BCA)=180o−(30o+x+40o+x)=110o−2x......(iii).(byanglesumpropertyoftriangles)From(ii)&(iii)110o−2x=90o−x⟹x=20o.∴∠AOC=90o−x=90o−20o=70o(fromii)Ans−OptionA.