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Offical GMAT Review 2nd Edition #162

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Offical GMAT Review 2nd Edition #162

by bookjive » Wed Feb 22, 2012 11:46 am
I'm not seeing a solution to this problem on the forum and the book's solution is hard to follow. Any ideas?

Here is the question without the picture:

In the figure above, point O is the center of the circle and OC=AC=AB. What is the value of x?
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by GMATGuruNY » Wed Feb 22, 2012 1:54 pm
bookjive wrote:I'm not seeing a solution to this problem on the forum and the book's solution is hard to follow. Any ideas?

Here is the question without the picture:

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
D) 32
E) 30
Image

The figure above shows the following relationships:
Since OC=AC, ∠AOB=∠OAC=x.
Since OA=OB, ∠OAB=∠ABC.
Since AC=AB, ∠ABC=∠ACB.
Thus, the following 3 angles are equal: ∠OAB=∠ABC=∠ACB.

We can plug in the answer choices, which represent the value of x.

Answer choice C: x=34
Image
The values in the figure above satisfy all of the conditions in the problem.
The sum of the angles inside ∆AOB = 34+34+34+68 = 170.
Since the sum of the angles inside a triangle must be 180, a greater answer choice is needed.
Eliminate C, D, and E.

Answer choice B: x=36
Image
The values in the figure above satisfy all of the conditions in the problem.
The sum of the angles inside ∆AOB = 36+36+36+72 = 180.
Success!

The correct answer is B.
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by pallasy » Sun Apr 01, 2012 4:00 pm
GMATGuruNY, how did you arrive at the conclusion that OA=OB based off OC = AC = AB?
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by GMATGuruNY » Sun Apr 01, 2012 4:54 pm
pallasy wrote:GMATGuruNY, how did you arrive at the conclusion that OA=OB based off OC = AC = AB?
Since OA and OB both radii of the circle, OA=OB.
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by mcdesty » Mon Jan 07, 2013 9:34 am
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by AdelinaWe » Tue Feb 03, 2015 2:19 pm
GMATGuruNY wrote:
pallasy wrote:GMATGuruNY, how did you arrive at the conclusion that OA=OB based off OC = AC = AB?
Since OA and OB both radii of the circle, OA=OB.
Dear Mitch,

I do not undestand why do you consinder the line AC as a angle bisector of < OAB ?
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by GMATGuruNY » Wed Feb 04, 2015 4:22 am
AdelinaWe wrote:
GMATGuruNY wrote:
pallasy wrote:GMATGuruNY, how did you arrive at the conclusion that OA=OB based off OC = AC = AB?
Since OA and OB both radii of the circle, OA=OB.
Dear Mitch,

I do not undestand why do you consinder the line AC as a angle bisector of < OAB ?
Answer choice C: x=34
Image
Here, ∠AOC=∠OAC=34.
Since the angles inside ∆AOC must sum to 180, ∠ACO = 180-34-34 = 112.

Image
Since angles that form a straight line must sum to 180, ∠ACO + ∠ACB = 180.
Thus, ∠ACB = 180-112 = 68.
Since ∠ACB = ∠ABC, ∠ABC = 68.

Image
Since the angles inside ∆ABC must sum to 180, ∠CAB = 180-68-68 = 34.

Result:
∠OAC = ∠CAB = 34, implying that AC bisects ∠OAB.
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by GMATGuruNY » Wed Feb 04, 2015 4:48 am
Alternate approach:

Image
∠ACB is an EXTERIOR ANGLE of ∆OAC.
An exterior angle is equal to the SUM OF THE TWO OPPOSITE INTERIOR ANGLES.
Thus, ∠ACB = x+y, as shown in the figure above.
GMATGuruNY wrote: 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
D) 32
E) 30
Image
Since OC=AC, ∠AOC = ∠OAC = x.

Image
Since exterior angle ∠ACB is equal to the sum of the two opposite interior angles,
∠ACB = ∠AOC + ∠OAC = x + x = 2x.
Since AC=AB, ∠ABC =∠ACB = 2x.

Image
Since radius OA = radius OB, ∠OAB = ∠ABC.
Thus, ∠OAB = ∠ABC = 2x.
Since ∠OAB = 2x and ∠OAC= x, ∠CAB = x.

Since the angles inside ∆ABC must sum to 180, we get:
x + 2x + 2x = 180
5x = 180
x = 36.

The correct answer is B.
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