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Are all angles of triangle ABC smaller than 90 degrees? (1) [#permalink]
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13 Nov 2008, 02:13
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Are all angles of triangle ABC smaller than 90 degrees?
(1) \(2AB = 3BC = 4AC\)
(2) \((AC)^2 + (AB)^2 > (BC)^2\)
Now, the OA to this question is A. First, I thought the answer is D, but I was wrong. But I also know that any right triangle has the equation \(a^2+b^2 = c^2\). In statement 2, we rather have \(a^2 + b^2 > c^2\) (assuming that we have the 3 sides to be a,b,c). How come that could also be a right triangle?



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 02:33
From the OA you post I got the solution
(1) 2AB = 3BC = 4AC
we have cos A = (AB^2 + AC^2  BC^2) / (2AB * AC)
from cosA, we know if A>0 or not, same with B, C => suff
(2) only know for angle A => insuff
=>D



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 02:44
B is insufficient as even in right angled triangle with right angle at B, stmt2 will hold true.



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 03:40
I know at least that A is correct because if it is a right triangle, then the ratio of the sides should be either 3:4:5, 5:12:13, etc...but from statement 1, the ratio of 2:3:4 is no way the ratio of a right triangle.
my main concern is with option B, because I honestly thought that it's suff. as "no", but according to the OA, statement 2 can mean both a right triangle and a nonright triangle...how??



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 03:50
tarek99 wrote: I know at least that A is correct because if it is a right triangle, then the ratio of the sides should be either 3:4:5, 5:12:13, etc...but from statement 1, the ratio of 2:3:4 is no way the ratio of a right triangle.
my main concern is with option B, because I honestly thought that it's suff. as "no", but according to the OA, statement 2 can mean both a right triangle and a nonright triangle...how?? Look at my earlier post. Draw a triangle with obtuse angle at B. Stmt2 will still hold true.



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 04:12
scthakur wrote: tarek99 wrote: I know at least that A is correct because if it is a right triangle, then the ratio of the sides should be either 3:4:5, 5:12:13, etc...but from statement 1, the ratio of 2:3:4 is no way the ratio of a right triangle.
my main concern is with option B, because I honestly thought that it's suff. as "no", but according to the OA, statement 2 can mean both a right triangle and a nonright triangle...how?? Look at my earlier post. Draw a triangle with obtuse angle at B. Stmt2 will still hold true. there's no way an obtuse will work. An obtuse angle is already more than 90 degrees. So if one of the angle is obtuse, there is simply no way that one of the remaining angles can even be 90. This is because if there is a second angle that has a 90 degrees, then the total sum of angles will be more than 180, which is impossible.



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Re: DS: Geometry [#permalink]
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13 Nov 2008, 04:13
Another way to think of this problem... A) 3 sides of the triangle always fix all angles. So we even haven't to go further in order to define presence of a >90 angle. B) Let's consider the case with huge AB and AC, and small BC. Let's ABC is near 0. Now, let's try to move in mind BC changing ABC angle from 0 to 180. It becomes obvious that we move from triangle with >90 angle at ABC~0 to triangle without >90 angle at AB=AC and again toward triangle with >90 angle at ABC~180. Moreover, duiring the movement we pass two right triangles (ABC=90 and ACB=90). Hope this help
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Re: DS: Geometry [#permalink]
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17 Nov 2008, 08:28
walker wrote: Another way to think of this problem...
A) 3 sides of the triangle always fix all angles. So we even haven't to go further in order to define presence of a >90 angle.
B) Let's consider the case with huge AB and AC, and small BC. Let's ABC is near 0. Now, let's try to move in mind BC changing ABC angle from 0 to 180. It becomes obvious that we move from triangle with >90 angle at ABC~0 to triangle without >90 angle at AB=AC and again toward triangle with >90 angle at ABC~180. Moreover, duiring the movement we pass two right triangles (ABC=90 and ACB=90).
Hope this help walker, your explanation is helpful. I agree with OA being A.










