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

#1**+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

#1**+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