Circle question?

1 Answer
P dilip_k
Apr 23, 2018

#OB# is radius and #ABQ# is tangent .

Hence #angleOBQ=90^@#

Here #OB=OC=OD="radius of the circle"#

#Delta ODC# is isosceles.

So #angleOCD=angleODC=x#

Similarly

#Delta OBC# is isosceles.

So #angleOCB=angleOBC=angleOBQ-2x=90-2x#

So #angle BCD=angle OCB+angle OCD=90-x#

So central #angle BOD=2angle BCD=180-2x#

Now
obtuse #angleBOD=180^@-y# as BODA is a cyclic quadrilateral .

Hence #180-y=180-2x#

#=>y=2x#

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