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Degree measure of angle? (GMAT on Saturday, please help!)

Expert replies
Source: — Problem Solving |

by Ad_Astra_Per_Aspera » Thu Jul 09, 2009 7:01 pm
let a be the measure of angle BAO. Since AB and OC have same length, they must both have the length of OB cuz OB and OC are the
radius. Since AB and OB have the same length this gives angle BOA = angle BAO = a

angle OBC + angle OBA = (a + a) + angle OBA because each side adds upto 180. So angle OBC = 2a

angle OCB = 2a as well sice OC and OB have the same length.

Property of triangle is that 3 angles must add up to 180 so we have: 2a + 2a + angle BOC = 180

angle BOC = 180 - 4a

Starting with statement 2. From the graph angle BCO = 2a and from this statement 2a = 40. so we can easily find a which is the degree measure of BAO. so statement 2 sufficient.

At this point we know that we can only have 1 of 3 answer choices right B or D or A.

Look at statement 1 next,

angle BOA + angle BOC + angle COD = 180

or a + 180 - 4a + 60 = 180 so we can find a which is the degree measure of angle BAO. so this statement is sufficient.

Hence D is the answer.
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by Adas » Thu Jul 09, 2009 7:36 pm
Are we allowed to assume that AC is a line segment?

I was hesitant to do this when solving the question because this is never specified...AB and BC could meet an angle other that 0 degrees....
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by Ad_Astra_Per_Aspera » Thu Jul 09, 2009 8:00 pm
assuming AC to be a line segment is the only way of successfully doing the question in my opinion. Unless the official answer is not D then that would surprise me a good deal. Please confirm the answer.
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by barnum » Thu Jul 09, 2009 9:42 pm
Adaś wrote:Are we allowed to assume that AC is a line segment?

I was hesitant to do this when solving the question because this is never specified...AB and BC could meet an angle other that 0 degrees....
This is a very important question. It is actually in the directions to the section that most people don't read. The GMAT specifically states that lines that appear to be straight can be assumed to be straight. So yes, you can assume AC is a straight line segment.

This is actually why other pictures are not drawn to scale. If the angle were to work out to say a minimal 3 degrees, the GMAT will exaggerate it to possibly 20 degrees to ensure that there is no question the line bends.
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by Adas » Fri Jul 10, 2009 3:37 am
The answer to the question is indeed D.

I must have missed those instructions (re: lines). Does this apply to both PS and DS questions?
barnum wrote:
Adaś wrote:Are we allowed to assume that AC is a line segment?

I was hesitant to do this when solving the question because this is never specified...AB and BC could meet an angle other that 0 degrees....
This is a very important question. It is actually in the directions to the section that most people don't read. The GMAT specifically states that lines that appear to be straight can be assumed to be straight. So yes, you can assume AC is a straight line segment.

This is actually why other pictures are not drawn to scale. If the angle were to work out to say a minimal 3 degrees, the GMAT will exaggerate it to possibly 20 degrees to ensure that there is no question the line bends.
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by barnum » Fri Jul 10, 2009 7:37 pm
Yes, it is in the directions for both problem solving and data sufficiency. If you have the OG you can read it in the front part (the diagnostic).
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Adaś wrote:Please help me solve this question and please show all your work. Your help is greatly appreciated :)
The key here is to look for isoceles triangles (always look for these in many such problems).

OB=OC, so let the two equal angles be X
AB =OB, so let the two equal angles be Z. Z is the angle of interest.

Use the Extended side of a triangle theorem to solve

1) 60+Y=180-X+Z (Extended side of ABO where the extension happens at O is 60+Y(the other angle of BOC) is equal to the two remote interior angles of ABO, Z and the complement of X)

And X =2Z( Extended side argument for ABO where extension happens at B with Angle OBC extended)
So u just need 2 equation in 2 unknowns Z and Y and you know u can solve. ( You don't need to solve)

Y+2X=180 .....1 ( Sum of angles in BOC) Substitute X=2Z in here you have Z and Y
Y+4Z=180

Do the same for the first equation and you have another equation in Y and Z. So You can solve for Z using these 2 equations. AGAIN YOU ARE NOT ASKED TO SOLVE.

2) This easier because
40= 2Z (40 HERE IS X IN THE FIRST PART SINCE TRIANGLE BOC IS ISOCELES)
Z=20

Choose D.
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Is x=z

by pink_08 » Thu Jul 30, 2009 7:24 pm
In this case will x =z ? since AB=OB=OC
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Re: Is x=z

by dtweah » Wed Aug 12, 2009 9:55 am
pink_08 wrote:In this case will x =z ? since AB=OB=OC
No.
Under 1) in the previous thread above
60+Y=180-X+Z
y+z=120

combine with y+4z=180

y+z=120
y+4z=180

Z=20

X=40

Under 2. X is given as 40. and Z=20
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