GMAT PREP DS
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Radius of semi circle is r say. OC=OB=OD=r. Therefore AB=r.
Statement 1 just gives the value of COB. All I can get is COA = 120 and CAO + ACO = 120. We need CAO (BAO). So, this is not enough.
Statement 2:
i) BCO = 40. This implies CBO is also 40 (OC = OB and isosceles triangle).
ii) from i, ABO = 140 (180-40). and BAO + BOA = 40.
iii) AB = OB (Given)
Hence BAO is equal to BOA. Hence BAO = 20.
Therefore statement2 itself is sufficient.
What is the OA?
Statement 1 just gives the value of COB. All I can get is COA = 120 and CAO + ACO = 120. We need CAO (BAO). So, this is not enough.
Statement 2:
i) BCO = 40. This implies CBO is also 40 (OC = OB and isosceles triangle).
ii) from i, ABO = 140 (180-40). and BAO + BOA = 40.
iii) AB = OB (Given)
Hence BAO is equal to BOA. Hence BAO = 20.
Therefore statement2 itself is sufficient.
What is the OA?
is OA d ?
1 ]
AB = OB = OC
now COD = 60
AOB + BOC + COD = 180
triangle BOC is isosceles triangle ... let's angle OBC = BCO = x
hence BOC = 180 - 2x
therefore our eqn becomes AOB + 180 - 2x + COD = 180
now ABO + OBC = 180
ABO + x = 180
even ABO is isosceles triangle .. let's say AOB = BAO = y
2y + ABO = 180
therefore 2y + 180 - x = 180
hence y = x / 2
putting y = x/2 in our previous equation
x/2 + 180 - 2x + 60 = 180
hence x = 40
therefore y = 20 ...i.e BAO = 20
2 ] already replied by jay2007
1 ]
AB = OB = OC
now COD = 60
AOB + BOC + COD = 180
triangle BOC is isosceles triangle ... let's angle OBC = BCO = x
hence BOC = 180 - 2x
therefore our eqn becomes AOB + 180 - 2x + COD = 180
now ABO + OBC = 180
ABO + x = 180
even ABO is isosceles triangle .. let's say AOB = BAO = y
2y + ABO = 180
therefore 2y + 180 - x = 180
hence y = x / 2
putting y = x/2 in our previous equation
x/2 + 180 - 2x + 60 = 180
hence x = 40
therefore y = 20 ...i.e BAO = 20
2 ] already replied by jay2007
Last edited by ashishrs on Wed Jul 25, 2007 4:38 am, edited 1 time in total.