Circle question?

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Answer 1 · 2018-04-23T01:01:48

$OB$ 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 $180-y=180-2x$

$=>y=2x$