Bunuel wrote:
PathFinder007 wrote:
HI Bunnel,
Could you please provide your comments on statement defined in question.
Thanks.
THEORY:Say the lengths of the sides of a triangle are a, b, and c, where the largest side is c.
For a right triangle: \(a^2 +b^2= c^2\).
For an acute (a triangle that has all angles less than 90°) triangle: \(a^2 +b^2>c^2\).
For an obtuse (a triangle that has an angle greater than 90°) triangle: \(a^2 +b^2<c^2\).
Points A, B and C form a triangle. Is ABC > 90 degrees?(1) AC = AB + BC - 0.001.
If AC=0.001, AB=0.001 and BC=0.001, then the triangle will be equilateral, thus each of its angles will be 60 degrees.
If AC=10, AB=5 and BC=5.001, then AC^2>AB^2+BC^2, which means that angle ABC will be more than 90 degrees.
Not sufficient.
(2) AC = AB --> triangle ABC is an isosceles triangle --> angles B and C are equal, which means that angle B cannot be greater than 90 degrees. Sufficient.
Answer: B.
Similar questions to practice:
http://gmatclub.com/forum/are-all-angle ... 29298.htmlhttp://gmatclub.com/forum/if-10-12-and- ... 90462.htmlHope it's clear.
Hi Bunuel, to find an obtuse angle within the constraints set by 1) I did the following (in bold). Is this approach okay?
1)
If AB + BC = 100, then angle ABC will be close to 180. This triangle is allowed because AC<AB+AC. I felt that this triangle allowed easier visualization of the obtuse angle.
And, as you stated if all sides = 0.001, then angle ABC will be 60.
2) Means that the triangle is isosceles and therefore has 2 equal angles.
2x+y=180
2x=180-y
Because y cannot be 0, x must be less than 90. Suff.