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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 non-right triangle...how??
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tarek99
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 non-right triangle...how??


Look at my earlier post. Draw a triangle with obtuse angle at B. Stmt2 will still hold true.
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tarek99
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 non-right 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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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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walker
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.



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