+0  
 
+2
230
1
avatar+814 

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? 

 Nov 24, 2018
 #1
avatar+107348 
+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

 

 

cool cool cool

 Nov 24, 2018

24 Online Users

avatar