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

We have our diagram , As :

Here we have joined AC and OB  ,

In $∆$ OBC

OC  =  OB                                      ( Radius of circle )

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

So, from angle sum property we get

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

30$°$ + 30$°$ +  $\angle$ BOC = 180$°$

$\angle$ 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,
$\angle$ BOC =  2 $\angle$ BAC , So

2$\angle$ BAC =  120$°$

$\angle$ BAC =  60$°$
And

As we know $\angle$ 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 )

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

Hence $∆$ ABE $\cong$ $∆$  ACE     ( By RHL rule )
So,
$\angle$ BAE  =  $\angle$ CAE                    ( From CPCT )

And
$\angle$ BAC  =  $\angle$ BAE  + $\angle$ CAE  , Substitute values , we get

x  + x  =  60$°$

2x  = 60$°$

x  =  30$°$                                         ( Ans )

And
$\angle$ COD  =  $\angle$ 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,
$\angle$ COD =  2 $\angle$ CBD , So

2y  = 30$°$

y  = 15$°$                                                                ( Ans )

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