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# In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d

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Director
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 09:49
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?

(1) Angle DAB = 45 degrees

(2) Angle ADC + angle BCD = 205 degrees

[Reveal] Spoiler:
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DS.jpg [ 17.79 KiB | Viewed 1620 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 10:44
I got C

Thanks
Director
Joined: 07 Jun 2004
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 12:51
The OA is D without explanation i think its a TYPO

can you let us know how you got C
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 13:21
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rxs0005 wrote:
The OA is D without explanation i think its a TYPO

can you let us know how you got C

No typo OA is correct.

Important property of polygons: Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides.

So for quadrilateral sum of the angles equals to $$180(4-2)=360$$ degrees.

Given: $$\angle{ABC}=110$$ and $$\angle{AED}=100$$. Question: $$\angle{BCD}-\angle{ADE}=?$$.

Now as $$\angle{ABC}=110$$ and $$\angle{AED}=100$$ and the sum of all angles is 360 degrees then the sum of other two angles in quadrilateral must be $$\angle{BCD}+\angle{BAE}=360-(110+100)=150$$.

(1) $$\angle{DAB}=45$$ --> $$\angle{BAE}=\angle{DAB}+\angle{DAE}=45+\angle{DAE}$$, but $$\angle{DAE}=180-100-\angle{ADE}=80-\angle{ADE}$$ (the sum of interior angles in triangle ADE equals to 180 degrees) --> $$\angle{BAE}=45+80-\angle{ADE}=125-\angle{ADE}$$. So $$\angle{BCD}+\angle{BAE}=150$$ becomes $$\angle{BCD}+125-\angle{ADE}=150$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.

(2) $$\angle{BCD}+\angle{ADC}=205$$ --> $$\angle{ADC}=180-\angle{ADE}$$ (straight line) --> $$\angle{BCD}+180-\angle{ADE}=205$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.

Hope it's clear.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 15:35
good Qn.
Thanks Bunuel.
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Director
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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13 Aug 2010, 15:40
thanks i now get it

i was taking interior angles as 180 which is not necessarily true
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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14 Aug 2017, 02:19
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d   [#permalink] 14 Aug 2017, 02:19
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