Offical GMAT Review 2nd Edition #162

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

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

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

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.

GMATGuruNY, how did you arrive at the conclusion that OA=OB based off OC = AC = AB?

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.

See attachment

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 ?

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

Here, ∠AOC=∠OAC=34.
Since the angles inside ∆AOC must sum to 180, ∠ACO = 180-34-34 = 112.


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


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

Result:
∠OAC = ∠CAB = 34, implying that AC bisects ∠OAB.

Alternate approach:


∠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

Since OC=AC, ∠AOC = ∠OAC = x.


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.


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.