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 \(BC\)

Also, should I use the power of point theorem?

mathtoo Nov 24, 2018

#1**+1 **

Angle CAD is an inscribed angle intersecting arc CD.....so..its measure is 1/2 of the arc = 30°

And since angle ABO = 60°, then angle BOA = 180 - 30 - 60 = 90°

So....we have 30 - 60 - 90 right triangle

And the side opposite the right angle = AB = 2BO = 10

And the side opposite the the 60° angle ABO = AO = 5√3 = √75 = the radius of the circle

Draw OB through to intersect the circle at M.....so OM is a radius = 5√3 = √75

So BM = OM - BO = [ 5√3 - 5] = [ √75 - 5 ]

And draw BO through to intersect the circle at N

So.....BN = [ 5√3 + 5] = [ √5 + 5 ]

An

So.....we can use the intersecting chord theorem to find BC

AB * BC = BM * BN

10 * BC = [ √75 - 5 ] * [ √75 + 5 ]

10 * BC = 75 - 25

10 * BC = 50 divide both sides by 10

BC = 50 / 10 = 5

CPhill Nov 24, 2018