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