Geometry DS Q (Iso Triangle)

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Geometry DS Q (Iso Triangle)

by milanproda » Tue Jan 04, 2011 2:04 pm
In an isosceles triangle DEF, what is the measure of angle EDF?

1- Angle DEF is 96 degrees

2- Angle DFE is 42 degrees

Answer: Statement 2 alone is sufficient, but state 1 alone is not sufficient.

I will try to explain my logic as to why I though it was D (Each statement alone is sufficient). If you are given one angle in an isosceles triangle, could you find every other angle measure in the triangle?

For example, say that in angle DEF, the main angle (aka the non congruent one) is 96 degrees, could you not split the triangle into two right triangles, split the 96 by 2 (48 degrees each) and with that discern the third angle? (48 plus 90= 138. 180-138= 42 degrees.

So couldnt the answer be 4 (or D)?
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by Tani » Tue Jan 04, 2011 2:15 pm
In an isosceles triangle two angles must be equal and one different.

1) If one angle is 96 degrees it cannot be one of the two equal angles (because you would have more than 180 degrees) so it must be the different angle and the others must be (180-96)/2 = 42 degrees.

2) If one angle is 42 degree we could have 42,42,96 (as seen in clue one), OR 42 could be the different angle and then the others would be (180-42)/2, giving us 69, 69, 42. Two possibilities = insufficient.

Answer A
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by milanproda » Tue Jan 04, 2011 2:21 pm
Thank you for the quick reply,

If stem one gives 96 degrees, and 96 can only be the non-equal angle, then I cannot assume that it is the largest angle? If 96 was split could you not find the other angles?
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by Tani » Tue Jan 04, 2011 2:27 pm
You are given 96 as an angle. You cannot split it, you have to have a 96 degree angle in the triangle. By splitting it you would have 2 triangles which is a different problem.
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by Geva@EconomistGMAT » Wed Jan 05, 2011 2:30 am
milanproda wrote:Thank you for the quick reply,

If stem one gives 96 degrees, and 96 can only be the non-equal angle, then I cannot assume that it is the largest angle? If 96 was split could you not find the other angles?
Would just like to point out that you don't need the split: if you know that the 96 is non-equal angle (which it must, since we cannot have a twin 96), then the other two angles must be equal. So the remaining 180-96 = 84 is split two ways between the two equal angles, and each is 42.

The problem with (2) is that you do not know whether the 42 angle is one of the equals (leaving the other equal angle as 42 and the last one as 96), or the non-equal angle (leaving the other two angles as half of 180-42=138, or 69). You don't need to split the given angle in order to find the other two: The minute you define your scenario (42 is one of the equals, or the third non-equal), you can find the other two angles without splitting the 42 one. But the problem remains that there are two scenarios: two different triangles that can be drawn, and the required angle DEF will be different for either scenario.
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