# O is the ci

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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