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

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

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New post 26 Dec 2013, 06:01
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A
B
C
D
E

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

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New post 26 Dec 2013, 06:20
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1
nechets wrote:
Image

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

Answer: C.

Hope it's clear.

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

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New post 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



Image

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

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New post 31 Jan 2018, 16:56
nechets wrote:
Image

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

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

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New post 24 Oct 2018, 02:22
1
nechets wrote:
Image

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

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New post 24 Oct 2018, 03:00
stne wrote:
nechets wrote:
Image

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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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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