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 $BC$.
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
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