VERBAL1 wrote:
The angles in a triangle are \(x\), \(3x\), and \(5x\) degrees. If \(a\), \(b\) and \(c\) are the lengths of the sides opposite to angles \(x\), \(3x\), and \(5x\) respectively, then which of the following must be true?
I. \(c \gt a+b\)
II. \(c:a:b=10:6:2\)
III. \(c^2 \gt a^2+b^2\)
A. I and III only
B. II and III only
C. I only
D. II only
E. III only
M08-31
Spoiler: OA
B
The OA currently in the original post is "B", which is not correct (as Bunuel's solution demonstrates).
The sum of any two sides in a triangle always exceeds the third, so that alone lets us rule out I and II. For II, you can also recall that the ratio of a triangle's angles is almost never the same as the ratio of the lengths of its sides; that's the whole reason the subject of trigonometry is not trivially easy.
III is true because, with angles x, 3x and 5x, using the fact that these sum to 180, we find x = 20. So the largest angle is 100 degrees. When one angle in a triangle is precisely 90 degrees, a^2 + b^2 = c^2 is true, where c is the longest side. When you make the largest angle bigger than 90 degrees, that creates more space opposite that angle, making the opposite side longer. So in any triangle with an angle bigger than 90 degrees, c^2 > a^2 + b^2 is true, where c is the longest side.
But with these answer choices, it's not even necessary to confirm that III works, since I and II are both false by the triangle inequalities.