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Re: M23-32 [#permalink]
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]
Expert Reply
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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Re: M23-32 [#permalink]
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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]
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]
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]
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]
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]
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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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.
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Re: M23-32 [#permalink]
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