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I think this requires some ingenuity.

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I think this requires some ingenuity.  [#permalink]

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New post 28 Jul 2003, 08:50
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I think this requires some ingenuity.

In the figure above, what is the value of c in terms of a and b?
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New post 28 Jul 2003, 09:52
c = 180┬║ - a - b. What were the choices?
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Shame on you [#permalink]

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New post 28 Jul 2003, 10:07
Buzz, easy choices never work with ETS :!:

Shame on you, but no offense :-D .

Have you looked at Princeton Review rule number one?
Never expect easy answer choices.
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New post 28 Jul 2003, 10:09
what were the choices? It will help to find out the correct solution if I see them.
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New post 28 Jul 2003, 10:14
Pls Let me know if,

c = 360 - a - A - B - 2b (assuming a,b is different from A,B)

If they are the same, then c = 360 - 2a - 3b
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New post 29 Jul 2003, 06:26
stolyar wrote:
I got c=360-3b-2a


Okay guys, I'm getting c=2a+b-180 (assuming A & B mean a & b)
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New post 29 Jul 2003, 07:12
Stolyar is right, but how? :-D
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New post 29 Jul 2003, 07:47
In any triangle, sum of the 3 angles is 180 degrees.

I have marked the 3 angles unknow as x, y and z

from the 3 triangles,

y = 180 - a - b
x = 180 - a - b
z = 180 - b - c

Now, x + y + z = 180 (because they together form a straight line which is 180 degrees).

so, 180 - a - b + 180 - a - b + 180 - b - c = 180
=> c = 360 - 2a - 3b
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hello [#permalink]

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New post 29 Jul 2003, 09:12
thanks, prakunda2000 :lol:
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New post 29 Jul 2003, 11:12
You are welcome curli05 :?
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another sol [#permalink]

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New post 29 Jul 2003, 12:53
the sum of the angles in a quad = 360

ie

2a + 3b +c = 360

=> c = 360 -2a-3b
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New post 01 Aug 2003, 09:20
Mr. HarshBaron,

Explain your quad way :shock:
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I think this requires some ingenuity.

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