Angles problem?

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Angles problem?

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Answer 1 · 2018-04-22T01:58:17

See below.

Explanation:

First note.

The tangent to a point on the circumference of a circle is always perpendicular to the radius at the same point.

From diagram:

$∠ O B Q = 90^{\circ}$ $/_OBQ=90^@$

Hence:

$∠ C B O = 90^{\circ} - 2 X$ $/_CBO=90^@-2X$

For triangle $C O B$ $COB$

$C O = B O$ $bb(CO)=bb(BO)$ radii of circle.

So:

$C O B$ $COB$ is isosceles.

Hence:

$∠ O C B = ∠ C B O = 90^{\circ} - 2 X$ $/_OCB=/_CBO=90^@-2X$

For triangle $D O C$ $DOC$

$C O = D O$ $bb(CO)=bb(DO)$ radii of circle.

$D O C$ $DOC$ is isosceles.

Hence:

$∠ D C O = ∠ C D O = X$ $/_DCO=/_CDO=X$

For $∠ B C D$ $/_BCD$

We know:

$∠ O C B = 90^{\circ} - 2 X$ $/_OCB=90^@-2X$

and

$∠ D C O = X$ $/_DCO=X$

Hence:

$∠ B C D = ∠ O C B + ∠ D C O = 90^{\circ} - 2 X + X = 90^{\circ} - X$ $/_BCD=/_OCB+/_DCO=90^@-2X+X=90^@-X$

$∠ C O B = 180^{\circ} - (∠ C B O + ∠ O C B)$ $/_COB=180^@-(/_CBO+/_OCB)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =180^@-(90^@-2X +90^@-2X)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =180^@-(180^@-4X)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =4X$

$/_DOC=180^@-(/_CDO+/_DCO)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =180^@-(X+X)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =180^@-(2X)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ =180^@-2X$

$/_DOB+/_DOC+/_COB=360^@$

$:.$

$/_DOB=360^@-/_DOC-/_COB$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \=360^@-(180^@-2X)-4X$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \=360^@-180^@+2X-4X$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \=180^@-2X$

$/_ODA = /_OBA=90^@$ Perpendicular.

$/_ODA+/_OBA+/_DOB+y=360^@$ Sum of angles in quadrilateral

$:.$

$90^@+90^@+(180^@-2X)+y=360^@$

$y=360^@-180^@-180^@+2X$

$y=2X$

Collecting everything:

$/_DCO=X$

$/_CBO=90^@-2X$

$/_OCB=90^@-2X$

$/_BCD=90^@-X$

$/_DOB=180^@-2X$

$y=2X$

I think there is probably a simpler method for this and maybe someone else will give a more concise way, but hope it helps.

Answer 2 · 2018-04-22T04:22:53

see explanation.

Explanation:

enter image source here
let $r$ be the radius of the circle,
As $OD=OC=r, => angleDCO=angleCDO=x$ ,
given $QBA$ is tangent to the circle and $angleCBQ=2x$ ,
$=> angleCBO=90-2x$ ,
as $OB=OC=r, => angleOCB=angleOBC=90-2x$ ,
$angleBCD=angleOCB+angleDCO=90-2x+x=90-x$ ,
As the angle at the center of the circle is twice the angle at the circumference
$angleDOB=2*angleBCD=2*(90-x)=180-2x$

As $ADOB$ is a quadrilateral, and the sum of the angles of a quadrilateral always adds up to $180^@$ ,
$=> y+angleADO+angleDOB+angleABO=360^@$ ,
$=> y=360-90-(180-2x)-90=360-360+2x=2x$

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Angles problem?

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