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M23-32

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M23-32  [#permalink]

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New post 16 Sep 2014, 01:20
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A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees

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Re M23-32  [#permalink]

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New post 16 Sep 2014, 01:20
Official Solution:

A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees

\(\angle OAB = \frac{180 - \angle BOA}{2} = \frac{(180 - \frac{360}{5})}{2} = 54\) degrees.

Answer: B
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M23-32  [#permalink]

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New post 14 Jul 2015, 05:45
Bunuel wrote:
Official Solution:

A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees

\(\angle OAB = \frac{180 - \angle BOA}{2} = \frac{(180 - \frac{360}{5})}{2} = 54\) degrees.

Answer: B


Hi bunuel,

I'd appreciate your help understanding this solution better.

To arrive at 360/5 for angle BOA, it looks like you've assumed the radii from each of the pentagon's vertices to the center divide the 360 degrees into 5 equal parts. Is this allowed? What if the pentagon had unequal sides. Wouldn't that vary the angles subtended at the center?

My geometry is not great. If I've made any incorrect assumptions above, please let me know.
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Re: M23-32  [#permalink]

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New post 14 Jul 2015, 09:02
samuraijack256 wrote:
Bunuel wrote:
Official Solution:

A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees

\(\angle OAB = \frac{180 - \angle BOA}{2} = \frac{(180 - \frac{360}{5})}{2} = 54\) degrees.

Answer: B


Hi bunuel,

I'd appreciate your help understanding this solution better.

To arrive at 360/5 for angle BOA, it looks like you've assumed the radii from each of the pentagon's vertices to the center divide the 360 degrees into 5 equal parts. Is this allowed? What if the pentagon had unequal sides. Wouldn't that vary the angles subtended at the center?

My geometry is not great. If I've made any incorrect assumptions above, please let me know.


Check the highlighted part above.

Regular polygon: a polygon with all sides and interior angles the same.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M23-32  [#permalink]

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New post 14 Feb 2016, 15:52
is it true for all REGULAR polygons inscribed in circle, any line from radii to each angle will split the angle in exactly half? That's the assumption I'm drawing from the explanation above..
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Re: M23-32  [#permalink]

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New post 11 Oct 2016, 01:57
Bunuel wrote:
samuraijack256 wrote:
Bunuel wrote:
Official Solution:

A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees

\(\angle OAB = \frac{180 - \angle BOA}{2} = \frac{(180 - \frac{360}{5})}{2} = 54\) degrees.

Answer: B


Hi Bunuel,

Would you be kind to help with a figure for the above question. I think it will help the CAT takers too. Thanks.
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Re: M23-32  [#permalink]

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New post 11 Oct 2016, 02:18
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Bunuel wrote:
A regular pentagon is inscribed in a circle. If \(A\) and \(B\) are adjacent vertices of the pentagon and \(O\) is the center of the circle, what is the value of \(\angle OAB\) ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees



The Q just asks you what each angle will be if it is HALVED...
Angles in Pentagon= (n-2)*180=(5-2)*180=540....
Each angle =540/5..
Half of it = 540/5 * 1/2=540/10=54...

When you inscribe a REGULAR polygon in a circle, the centre will be same for both polygon and circle..
Also almost all times, when you join the side wth centre , it will bisect the angle..
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Re: M23-32  [#permalink]

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New post 05 Nov 2016, 08:23
Question: What about the rules for inscribed angles? The vertices of the pentagon divide the circle into 5 equal parts. So 360/5 = 72

So if angle AOB = 72, angles OAB=ABO as it is a regular pentagon. So 2x + 72 = 90, x = 9. This is obviously not correct. What am I doing wrong?
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Re: M23-32  [#permalink]

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New post 29 Mar 2018, 09:54
The angle at the center is 360/5 = 72. 180-72=108/2 = 54. The answer is option B
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Re: M23-32  [#permalink]

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New post 29 Mar 2018, 12:04
Is someone able to provide a diagram for this? I am having trouble visualizing. I came up with 72 as the answer but clearly am missing a step.
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Re: M23-32  [#permalink]

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New post 29 Mar 2018, 12:39
I'm not sure why we are subtractig from 180, can someone explain?
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Re: M23-32  [#permalink]

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New post 09 Apr 2018, 15:14
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I haven't made 5 posts yet so I can't attach the image. I will (try to remember to) edit this when I have.
The question stem just explains that the figure is a regular pentagon divided into congruent(?) isosceles triangles. They are isosceles as the sides are radii from the circumcentre to the circumcircle.
The three angles are: ∠AOB + two equal bottom angles. The question is not asking for ∠AOB (360/5), it is asking for ∠ABO.

The formula in Bunuel's second post is 180º less the known angle then divided by 2.
Re: M23-32 &nbs [#permalink] 09 Apr 2018, 15:14
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