b2bt wrote:
Bunuel wrote:
Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient.
(2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A
Is it right to say that the triangle is acute (and not right triangle) if \(BC^2 < AC^2 + AB^2\) and BC is the longest side of the triangle ?
Also, I did not understand Statement (1). How can we get the angles from sides? Is there a formula?
If the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then:
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: \(a^2 +b^2<c^2\).
As for your other question: you don't need to know any formula. It's enough to know that you CAN find the angles because the triangle will be fixed.
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