I) AC =3x, BC = 4x, AB = 6x. 3x²+4x²=5x²
AC² + BC² < AB² if it was = AB² then angle C =90°
As the ratio are given it is possible to say angle C is > 90°
Sufficient.
II) AC²+AB²> BC²
not sufficient as the ratio are not given. Insufficient
IMO A
ABC triangle
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bpgen
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Thanks all! Answer would be A alone sufficient. Here is bit detail explanation:
As we know cosine formulas, and cosX always >0 if it's angle <90 degree.
cosA=(-a^2+b^2+c^2)/(2bc)
cosB=(a^2-b^2+c^2)/(2ac)
cosC=(a^2+b^2-c^2)/(2ab)
i.e for acute angled triangle, (-a^2+b^2+c^2) ,(a^2-b^2+c^2) and (a^2+b^2-c^2) should be >0
Therefore, if you take ratio of sides: AC =3x, BC = 4x, AB = 6x
we have:
9+16>36
9+36>16
36+16>9
Hence triangle is acute angled triangle(all angles of triangle ABC smaller than 90 degrees).
As we know cosine formulas, and cosX always >0 if it's angle <90 degree.
cosA=(-a^2+b^2+c^2)/(2bc)
cosB=(a^2-b^2+c^2)/(2ac)
cosC=(a^2+b^2-c^2)/(2ab)
i.e for acute angled triangle, (-a^2+b^2+c^2) ,(a^2-b^2+c^2) and (a^2+b^2-c^2) should be >0
Therefore, if you take ratio of sides: AC =3x, BC = 4x, AB = 6x
we have:
9+16>36
9+36>16
36+16>9
Hence triangle is acute angled triangle(all angles of triangle ABC smaller than 90 degrees).
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kstv
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In GMAT we try to reason just with the basics.
If one knows the cosines formula very good for him/her as it will help them to solve itquickly .
But I was just thinkin of it is this fashion so as not to feel discouraged if I am unable to recall the formula
I) AC =3x, BC = 4x, AB = 6x.
3x²+4x²=5x²
when AC and BC are 3 and 4 respectively and AB = 5 angle ACB = 90°
the slightest enlongation of AC will make AC and BC diverge or angle ACB will be little > 90°
conversely contraction of AC will make AC and BC converge or angle ACB will be little < 90°
Even at this stage Option A is fine but
angle ACB will be the largest angle as it is opposite the longest side.
It value of ACB whether it is = < or > 90° will surely help decide the other.
Still for more clarity since AB = 6 angle ACB > 90°
angle CAB + ABC < 90°. Individually they have to less than 90°.
If one knows the cosines formula very good for him/her as it will help them to solve itquickly .
But I was just thinkin of it is this fashion so as not to feel discouraged if I am unable to recall the formula
I) AC =3x, BC = 4x, AB = 6x.
3x²+4x²=5x²
when AC and BC are 3 and 4 respectively and AB = 5 angle ACB = 90°
the slightest enlongation of AC will make AC and BC diverge or angle ACB will be little > 90°
conversely contraction of AC will make AC and BC converge or angle ACB will be little < 90°
Even at this stage Option A is fine but
angle ACB will be the largest angle as it is opposite the longest side.
It value of ACB whether it is = < or > 90° will surely help decide the other.
Still for more clarity since AB = 6 angle ACB > 90°
angle CAB + ABC < 90°. Individually they have to less than 90°.
I think both are sufficient
the first one is obvious
the second one:
If angle BAC = 90 then AC^2 + AB^2 = BC^2
The only way to change the ratio, without changing the lengths, is to modify the angle BAC...
If you increase the angle above 90, then AC^2 + AB^2 < BC^2
If you decrease the angle below 90, then AC^2 + AB^2 > BC^2
Hence the info is sufficient (all angles are <90)
the first one is obvious
the second one:
If angle BAC = 90 then AC^2 + AB^2 = BC^2
The only way to change the ratio, without changing the lengths, is to modify the angle BAC...
If you increase the angle above 90, then AC^2 + AB^2 < BC^2
If you decrease the angle below 90, then AC^2 + AB^2 > BC^2
Hence the info is sufficient (all angles are <90)
