Question

explain this plz .............

Answer 1

As OB=BC,
Angle BOC = Angle BCO = y (angles opposite to equal sides are equal)

Therefore,
angle OBC =180°-angle BOC -angleBCO
=180°-y-y
=180°-2y

AsABC is a straight line,
therefore
angle ABO+ angle CBO=180°
angle ABO+180°-2y=180°
AngleABO=2y

As OA and OB are radii of the same circle, therefore OA=OB and thus
Angle ABO= Angle OAB=2y

As x is the exterior angle of triangle OAC,
therefore angleOAC+angleOCA=x

y+2y=x

or 3y=x