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.

Answer A.

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PhD in Applied Mathematics

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