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# I think this requires some ingenuity.

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I think this requires some ingenuity.  [#permalink]  28 Jul 2003, 07:50
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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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c = 180┬║ - a - b. What were the choices?
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Shame on you [#permalink]  28 Jul 2003, 09:07
Buzz, easy choices never work with ETS

Shame on you, but no offense .

Have you looked at Princeton Review rule number one?
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what were the choices? It will help to find out the correct solution if I see them.
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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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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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Stolyar is right, but how?
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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]  29 Jul 2003, 08:12
thanks, prakunda2000
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You are welcome curli05
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another sol [#permalink]  29 Jul 2003, 11:53
the sum of the angles in a quad = 360

ie

2a + 3b +c = 360

=> c = 360 -2a-3b
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Mr. HarshBaron,

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