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# geometry

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Senior Manager
Joined: 12 Mar 2009
Posts: 308

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20 Jun 2009, 03:29
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Kudos [?]: 454 [0], given: 1

Manager
Joined: 30 May 2009
Posts: 214

Kudos [?]: 133 [1], given: 0

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20 Jun 2009, 15:26
1
KUDOS
(1) We are given COD = 60.

Let angle BAO = x.
Let angle OBC = y.
Let Angle BOC = z.

Now consider the triangle OBC.
sum of all angles of a triangle = BOC + OBC + BCO = 180
=> z + y + y = 180 (Because OBC = OCB = y since OB = OC = radius of the circle)
=> z + 2y = 180
=> z = 180 - 2y

Now conside the straight line AOD. Straingle angle formed at point O = 180
=> AOB + BOC + COD = 180
=> x + z + 60 = 180 (becaue AOB = BAO = x since AB=BO=CO=radius of the circle)
=> z = 120 - x

Also consider the triangle ABO. Exterior angle OBC = sum of interior opposite angles = BAO + AOB
=> y = 2x

Using all 3 euqations above, we can see that BAO = x = 20.
Hence SUFF

(2) Given that Angle BCO = 40.
Similar to step 1 we can derive that BAO(x) = y/2 = BAO/2 = 20
Hence SUFF

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Manager
Joined: 12 Apr 2006
Posts: 213

Kudos [?]: 30 [0], given: 17

Location: India

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24 Jun 2009, 03:42
sdrandom1

+1 for explanation.

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Re: geometry   [#permalink] 24 Jun 2009, 03:42
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