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In the figure above, point O is the center of the circle and

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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 12 Oct 2017, 11:56
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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In the figure above, point O is the center of the circle and OC = AC = AB. What is the value of x ?

(A) 40
(B) 36
(C) 34
(0) 32
(E) 30


Since OC = AC, ∆AOC is an isosceles triangle, which means ∠OAC is also x°
Image

Since all 3 angles in ∆AOC must add to 180°, we can conclude that ∠OCA = (180-2x)°
Image

Since angles on a LINE must add to 180°, we can conclude that ∠ACB = 2x°
Image

Since AC = AB, ∆ACB is an isosceles triangle, which means ∠CBA is also 2x°
Image

Finally, since OA and OB are radii of the same circle, we know that ∆OAB is an isosceles triangle, which means ∠OABis also 2x°
Image

At this point, we can see that the 3 angles ∆OAB are x°, 2x° and 2x°
Since the angles in a triangle must add to 180°, we can write: x° + 2x° + 2x° = 180°
Simplify: 5x = 180
Solve: x = 36

Answer: B

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In the figure above, point O is the center of the circle and [#permalink]

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New post 20 Oct 2017, 17:58
[quote="VeritasPrepKarishma"]
Hi,

if OC = AC = AB, then how come the three angles are not equal? (it's illustrated as x, y, and y in your diagram versus all y's)

Thanks

Kudos [?]: 0 [0], given: 98

In the figure above, point O is the center of the circle and   [#permalink] 20 Oct 2017, 17:58

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In the figure above, point O is the center of the circle and

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