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

Guest May 17, 2018

Best Answer 

 #1
avatar+7054 
+2

 

Because an inscribed angle is half the angle measure of its intercepted arc,

m∠CAD  =  (1/2)( m arc CD )

m∠CAD  =  (1/2)( 60° )

m∠CAD  30°

 

OA and OC are both radii of circle O, so they are the same length. So △AOC is isosceles, and

m∠OCA  =  m∠CAD 

m∠OCA  =  30°

 

Since  ∠OBC  and  ∠OBA  form a straight line,

m∠OBC  +  m∠OBA  =  180°

m∠OBC  =  180° - m∠OBA

m∠OBC  =  180° - 60°

m∠OBC  =  120°

 

Since the sum of the angles in a triangle is  180° ,

m∠BOC  +  m∠OCA  +  m∠OBC  =  180°

m∠BOC  +  30°  +  120°  =  180°

m∠BOC  =  180° - 120° - 30°

m∠BOC  30°

 

Since  m∠BOC m∠OCA ,  triangle OBC  is isosceles, and

the length of BC  =  the length of BO

the length of BC  =  5

 
hectictar  May 17, 2018
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1+0 Answers

 #1
avatar+7054 
+2
Best Answer

 

Because an inscribed angle is half the angle measure of its intercepted arc,

m∠CAD  =  (1/2)( m arc CD )

m∠CAD  =  (1/2)( 60° )

m∠CAD  30°

 

OA and OC are both radii of circle O, so they are the same length. So △AOC is isosceles, and

m∠OCA  =  m∠CAD 

m∠OCA  =  30°

 

Since  ∠OBC  and  ∠OBA  form a straight line,

m∠OBC  +  m∠OBA  =  180°

m∠OBC  =  180° - m∠OBA

m∠OBC  =  180° - 60°

m∠OBC  =  120°

 

Since the sum of the angles in a triangle is  180° ,

m∠BOC  +  m∠OCA  +  m∠OBC  =  180°

m∠BOC  +  30°  +  120°  =  180°

m∠BOC  =  180° - 120° - 30°

m∠BOC  30°

 

Since  m∠BOC m∠OCA ,  triangle OBC  is isosceles, and

the length of BC  =  the length of BO

the length of BC  =  5

 
hectictar  May 17, 2018

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