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# In the figure above, the vertices of triangle ABC lie on the circumfer

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Joined: 04 Oct 2013
Posts: 71
Location: Brazil
GMAT 1: 660 Q45 V35
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In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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26 Dec 2013, 06:01
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Difficulty:

25% (medium)

Question Stats:

78% (01:21) correct 22% (01:31) wrong based on 141 sessions

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In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?
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GC_Circles-e1517384956903.png [ 10.87 KiB | Viewed 456 times ]
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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26 Dec 2013, 06:20
1
1
nechets wrote:

In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?

All angels:

Notice that OB = OC = radius, thus triangle OBC is isosceles. Therefore angle OCB = angle OBC --> x + x + 70° = 180° --> x = 55°.

Hope it's clear.

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Imagem1.jpg [ 27.04 KiB | Viewed 2394 times ]

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Re: In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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31 Jan 2018, 00:50
nechets wrote:

In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?

Solution

• Assuming the O is the center of the circle and AC is the diameter.
o We can infer that the lines OB = OC = OA = radius of the circle.
• Let us focus on the triangle OBC, in which OB = OC.
o Since, it is an isosceles triangle
o Angle OBC = Angle OCB = x
o And we know that angle BOC = 70
• Per our conceptual understanding –
o Sum of internal angles of a triangle = 180
o Thus, for triangle OBC, we can write
 70 + x + x = 180
 2x = 110
 x = 55
• Thus, the correct answer is Option C.

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Saquib
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Re: In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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31 Jan 2018, 16:56
nechets wrote:

In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

Since sides OC and OB are radii, side OC = side OB, creating an isosceles triangle, and thus angle OCB = angle OBC = x.

Thus, we have:

2x + 70 = 180

2x = 110

x = 55

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Re: In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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24 Oct 2018, 02:22
1
nechets wrote:

In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?

Dear Moderator,
The figure of the above question is not visible , hope you will look into this. Thank you.
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- Stne
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: In the figure above, the vertices of triangle ABC lie on the circumfer  [#permalink]

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24 Oct 2018, 03:00
stne wrote:
nechets wrote:

In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?

(A) 35
(B) 45
(C) 55
(D) 65
(E) 70

I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?

Dear Moderator,
The figure of the above question is not visible , hope you will look into this. Thank you.

__________________
Fixed that. Thank you.
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Re: In the figure above, the vertices of triangle ABC lie on the circumfer   [#permalink] 24 Oct 2018, 03:00
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