Answer :


Since it is given that BOC = 100° and by the theorem:-


The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.



In ΔABC, as AB = BC, and we know that angle opposite to equal sides are equal.


ACB = ABC = t


As the sum of all angles in a triangle is equal to 180°


2t + BAC = 180


2t = 180 – 50


t = 65°


ABC = 65°


Similarly, in ΔBOC, BO = OC, applying the same principle as above, we get,


2OBC + BOC = 180


2OBC = 180 – 100


OBC = 40°


Also, ABO = ABC - OBC


ABO = 65 – 40 = 15°


The values of ABC and ABO are 65° and 15°.


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