In the given figure. O is the centre of circle,angle BCO=30 angle AEB=90 and OD || BC find x and y.

Answer :

We have our diagram , As :

Here we have joined AC and OB  ,

In OBC

OC  =  OB                                      ( Radius of circle )

OBC  = OCB = 30°    ( From base angle theorem , As OC  =  OB )

So, from angle sum property we get

OBC +  OCB +  BOC = 180°  , Substitute values , we get

30° + 30° +   BOC = 180°

  BOC = 120°
And
we know from theorem " The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.  "
So,
BOC =  2 BAC , So

2 BAC =  120°

BAC =  60°
And

As we know AEB  =  90° , So OE perpendicular on BC , and we know a line from center to any chord is perpendicular than that line also bisect chord )
SO,
CE  =  BE                            ---- ( 1 )

In ABE and   ACE

CE  =  BE                              (  From equation )

AEB  =  AEC  =  90°     ( Given )
And
AE  =  AE                          ( Common side )

Hence ABE   ACE     ( By RHL rule )
So,
BAE  =  CAE                    ( From CPCT )

And
BAC  =  BAE  + CAE  , Substitute values , we get

x  + x  =  60°

2x  = 60°

x  =  30°                                         ( Ans )

And
COD  =  OCB      = 30°                          ( Alternate interior angles As given OD  | | BC and take OC as transversal line )

And

we know from theorem " The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.  "
So,
COD =  2 CBD , So

2y  = 30°    

y  = 15°                                                                ( Ans )



 

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