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In the figure shown, point O is the center of the semicircle and B, C, [#permalink]

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29 Dec 2008, 09:25

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

43% (02:46) correct
57% (01:51) wrong based on 163 sessions

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In the figure shown, point O is the center of the semicircle and B, C, D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO ?

(1) The degree measure of angle COD is 60º. (2) The degree measure of angle BCO is 40º.

Re: In the figure shown, point O is the center of the semicircle and B, C, [#permalink]

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31 Dec 2008, 02:03

D

from the question we can know,

AB = OC ---1) BO = OC ---2) since both are radius. it makes triangle BOC and triangle BAO isoseles.

lets take BAO = BOA = x ----- 3) CBO = BCO = a ------ 4) BOC = y ABO = z

Stmt1: COD is 60

so we can say x+y = 120 -------(5) (from the line AD) also, z+a = 180 --------------(6) from triangle BAO, 2x+z = 180 --------------(7) from triangle BOC, 2a +y = 180 ---------------(8)

solving (6), (7), and (8) we get, 4x+y = 180 ---- (9)

from (5) and (9) we get x = 20 which is BAO --- SO SUFF.

stmt 2: this gives more close clue that a= 40 by using isoles triangle propety we get x = 20. SUFF

Re: In the figure shown, point O is the center of the semicircle and B, C, [#permalink]

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26 May 2009, 23:30

I have a question: How can you assume ABC is a straight line?

That's breaking the most fundamental GMAT rule-- assuming a shape is drawn properly!

I did this question and came to C, I know the answer is D, but how can you just assume that ABC is straight and hence we can sum up the angles to 180 on it?

See my attached schematic... if we don't assume ABC is straight, then the answer is C as my diagram shows that the shapes change.

Re: In the figure shown, point O is the center of the semicircle and B, C, [#permalink]

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22 Oct 2009, 07:52

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1st step: Write down data and assign variables: BAO = BOA = x CBO = BCO = a BOC = y ABO = z COD = b

2nd step: Write down the possible equations: 1) Triangle BAO: 180=z+2x 2) Triangle BOC: 180=y+2a 3) Line AOD: 180=x+y+b 4) Line ABC: 180=z+a -> 4 equations & 5 variables.

Stat1: gives the value of b -> 4 equations & 4 variables -> we can solve for x: sufficient

Stat2: gives the value of a -> 4 equations & 4 variables (or 3 & 3, as equation 3 becomes unnecesary) -> we can solve for x: sufficient

Now, if as Hades said, ABC is not a line, then equation 4 dissapears, and we need both a and b to solve for x -> answer would be C. But this is an official gmat prep problem, so I guess the OA must be right, and OA is D.

Could somebody help determine how could the answer be D without the assumption that ABC is a line? _________________

Re: In the figure shown, point O is the center of the semicircle and B, C, [#permalink]

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02 Oct 2014, 00:37

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In the figure shown, point O is the center of the semicircle and B, C, D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO ?

(1) The degree measure of angle COD is 60º. (2) The degree measure of angle BCO is 40º.

In the figure shown, point O is the center of the semicircle and B, C, D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO ?

(1) The degree measure of angle COD is 60º. (2) The degree measure of angle BCD is 40º.

Write down everything you know from the stem:

BO=CO=r=AB --> BOC and ABO are isosceles. <BAO=<BOA and <BCO=<OBC <CBO=2<BAO

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