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# M17 # 04

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Manager
Joined: 18 Oct 2008
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25 Jul 2010, 07:44
Are all angles of triangle $$ABC$$ smaller than 90 degrees?

1. $$2AB = 3BC = 4AC$$
2. $$AC^2 + AB^2 \gt BC^2$$

[Reveal] Spoiler:
Statement (1) by itself is sufficient. S1 completely defines the shape of the triangle, although not its size. Knowing the shape of the triangle, we are able to answer the question.

Statement (2) by itself is insufficient. The inequality holds in an acute-angled triangle (the answer is "yes") but it also holds if angle $$ABC$$ is right (the answer is "no").

Any alternate explanations ?
Manager
Joined: 16 Apr 2010
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25 Jul 2010, 10:20
Hi,

Regarding statement 1, the angles of a triangle are proportional to the length of the sides.
Thus the angles are in the ratio of 2,3 and 4 and are thus all less than 90 degrees.

As for statement 2, note the below:
- When the sum of the squares on two sides of a triangle is greater than the square on the third side, the same holds good for other pairs of sides as well, and the triangle is acute .
- When the sum of the squares on two sides of a triangle is equal to the square on the third side, then the triangle is right angled .
- When the sum of the squares on two sides of a triangle is less than the square on the third side, then the triangle is obtuse angled .

Each statement alone is sufficient to answer the question...

Regards,
Jack
Intern
Joined: 24 Jun 2010
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Location: Toronto
Schools: Berkeley Haas, UCLA Anderson, NYU Stern
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28 Jul 2010, 13:35
Statment 2 would need to be changed so that
side 1 ^2 + side 2 ^2 >= side 3^2

As jakolik mentioned that if the square of both smaller sides is larger than the square of the longest side, then the triangle would be acute, and the statment gives enough info on its own to answer.

I also think that each statement alone is sufficient
Senior Manager
Joined: 24 Mar 2011
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Location: Texas
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16 Jul 2011, 14:03

2nd statement of the question, holds true i think for the question to be correct. So it should be D.
However, A is the official answer.

So either the statement 2 needs to be changed to 'greater than equal to' for the OA to hold true. OR the OA needs to be corrected.
Intern
Joined: 06 Feb 2011
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09 Aug 2011, 21:44
2nd statement says us that the sum of squares of 2 sides (side AB and AC) is greater than the third side. This statement implies only and only that the angle contained by AB and AC is acute, i.e angle BAC is acute. Since this angle is acute, one of the other 2 angles can be greater than 90 (as in 30-30-120) or less than 90 (60-60-60) or equal to 90 (30-60-90). Thus this statement is insufficient as we are not sure whether the other angle is greater than or less than 90.

Try these triangles - AB=2, AC=root 3, BC=1 (hint: its a 30-60-90 triangle) and AB=AC=BC=5. Hope it helps.
Intern
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01 Dec 2011, 14:27
Here i think is a general rule

For any triangle with side a, b, c
a^2 = b^2 + c^2 - 2bc Cos(t) where t is the angle between the sides b and c

If t = 90, this is a right angled triangle and the rule becomes the pythagorus theorem.
If t < 90, then cos t > 0, => a^2 < b^2 + C^2

For a triangle to be acute, all three angles must be < 90
So the above rule must be satisfied for all three sides.

In other words, a triangle is acute if
a^2 < b^2 + C^2, b^2 < a^2 + c^2, and c^2 < a^2 + b^2

Condition 1 - does not satisfy the above condition, so we know that all three angles are not < 90

Condition 2 - satisfies one of the condition, but we do not know about the other 2. - Insufficient
Manager
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28 Jul 2012, 02:19
Ok so in statement 1; a unique ratio tells us that shape is fixed and hence angles are fixed but can we actually determine if it's an acute or an obtuse triangle; I mean do we need to determine the actual angles from the given ratio to prove sufficiency.

Also, using the ratio of the sides as given in statement 1, how can we determine the actual angles?
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28 Jul 2012, 03:00
1
This post was
BOOKMARKED
jakolik wrote:
Hi,

Regarding statement 1, the angles of a triangle are proportional to the length of the sides.
Thus the angles are in the ratio of 2,3 and 4 and are thus all less than 90 degrees.

As for statement 2, note the below:
- When the sum of the squares on two sides of a triangle is greater than the square on the third side, the same holds good for other pairs of sides as well, and the triangle is acute .
- When the sum of the squares on two sides of a triangle is equal to the square on the third side, then the triangle is right angled .
- When the sum of the squares on two sides of a triangle is less than the square on the third side, then the triangle is obtuse angled .

Each statement alone is sufficient to answer the question...

Regards,
Jack

Regarding statement 1, "the angles of a triangle are proportional to the length of the sides."

This is definitely not true! There is no such theorem.

The so-called "sinus theorem" states that for any triangle, the following equality holds:

$$\frac{AB}{sinC}=\frac{BC}{sinA}=\frac{AC}{sinB}=2R$$,
where $$A, B, C$$denote the angles of the triangle and $$R$$ is the radius of the circumscribed circle. $$sin$$ is the $$sinus$$ function, and the value of $$sina$$ is not proportional to the value of $$a$$.

From statement (1) we can deduce that $$AB = 6x, BC = 4x, AC = 3x$$, for certain positive $$x$$. Then $$AB^2=36x^2>BC^2+AC^2=25x^2$$, which means the triangle is obtuse angled, with angle C greater than 90. Therefore (1) is sufficient.

(2) is not sufficient, as it only proves that angle A is not obtuse, but nothing is known about the other two angles.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Re: M17 # 04   [#permalink] 28 Jul 2012, 03:00
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# M17 # 04

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