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In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO = 5$, and angle ABO = arc CD = 60 degrees. Find the length of $BC$.

 

 

 

 Jun 29, 2018
 #1
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If arc CD   = 60°, the angle CAD is  1/2 of this  since it  intercepts this arc

So CAD  = 30°  = BAO

And since BAO  and ABO  =  30  + 60  = 90°

Then,in triangle BAO,  angle BAO  must =  90°  = angle  BOD

 

So....triangle  BAO  is a 30-60-90 right triangle

 

Now....draw  OC....and it will be  equal  to   OD  since both are radii

 

And if arc CD  = 60°....then angle central angle COD  = 60°

 

And since angle BOD   = 90 °

Then angle BOC  =  angle BOD  - angle COD  = 90 - 60  = 30°  

And angle ABO  = 60  so angle CBO  is supplemental  = 120°

So  angle   BCO  =  180  - angle BOC - angle CBO  =  180 - 30 - 120 = 30°

 

Thus triangle BOC  is isosceles with  angle BOC  = angle BCO

But BO  =  5....so.....BC  has the same measure  = 5

 

 

 

cool cool cool

 Jun 29, 2018
edited by CPhill  Jun 29, 2018

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