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# Geometry Problem

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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 $$BC$$

Also, should I use the power of point theorem?

mathtoo  Nov 24, 2018
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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