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 )
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 )