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GMATPrep DS: Triangles and Angles

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GMATPrep DS: Triangles and Angles [#permalink]

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New post 17 Aug 2008, 08:48
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I got this question right, but just by thinking about the relationships of the different points on the triangle. Afterwards, I did this mathematically, and it took me a very long time with a large system of equations. Anyone have any thoughts on a quicker but mathematical way of doing this?
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Last edited by zonk on 17 Aug 2008, 09:49, edited 2 times in total.
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Re: GMATPrep DS: Triangles and Angles [#permalink]

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New post 17 Aug 2008, 09:15
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Say angle PTR=x. Then angle PRT= 90-x (for right triangle PRT)

1) triangle QRS is isoceles. so angle rsq = angle rqs = (180-(90-x))/2 = 90/2 + x/2
Insufficient. for dtermining x as angle tsu is still unknown.

2) similarly will give you tus=tsu = (180-x)/2 = 180/2-x/2
Insufficient as with this alone rsq is unknown

Combine to get x= 180-rsq-tsu

So C
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Re: GMATPrep DS: Triangles and Angles [#permalink]

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New post 17 Aug 2008, 09:14
Unsure if my solution is right, give your thoughts.

Angle R + Angle T = Angle P = 90 degrees.

Angle R + 2A = 180 degrees (2A because QR = RS)
Angle T + 2B = 180 degrees (2B because ST = TU)

Angle R + 2A + Angle T + 2B = 360 degrees (1 triangle = 180 degrees)
90 degrees + 2A + 2B = 360 degrees
2A + 2B = 270 degrees

A + x + B = 180 degrees (given it's a straight line on the triangle)

Simplify 2A + 2B = 270 degrees by dividing by 2, which results: A + B = 135 degrees.

A + B + x = 180 degrees
135 degrees + x = 180 degrees
x = 45 degrees regardless what value A and B are as long as the given conditions are met.

Therefore, an answer can be found when both conditions are met.
Re: GMATPrep DS: Triangles and Angles   [#permalink] 17 Aug 2008, 09:14
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GMATPrep DS: Triangles and Angles

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