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Re M0831
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15 Sep 2014, 23:38
Official Solution: 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 According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (\(c\)) cannot be larger than the sum of the other two sides (\(a\) and \(b\)). Note that II is basically the same as I: if \(c=10\), \(a=6\) and \(b=2\) then \(c \gt a+b\), which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them). Now, if interested why III is true: as the angles in a triangle are \(x\), \(3x\), and \(5x\) degrees then \(x+3x+5x=180\). Hence \(x=20\), \(3x=60\), and \(5x=100\). Next, if angle opposite side \(c\) were 90 degrees, then according to Pythagoras theorem \(c^2=a^2+b^2\), but since the angle opposite side \(c\) is more than 90 degrees (100) then side \(c\) is larger, hence \(c^2>a^2+b^2\). Answer: E
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Re: M0831
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18 Feb 2016, 01:29
Since C>ab but not a+b, (1) is wrong. a+b>c on the other hand. (2) is about ratio of length. Though it might seem that we cannot find length of arms just from angel degrees, but it is ratio not the actual length. So we find the (1) from (2) in fact. So, don't hold. (3) is right since c is even greater that right angel.



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Re: M0831
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20 Feb 2016, 13:22
Bunuel 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 we can eliminate right away A and C. in a triangle, the sum of the 2 sides will ALWAYS be greater than the third side. we are left with B, D, and E. since we know the angles are in x, 3x, and 5x ratio, the sides must be in the same ratio. c:a:b=5:1:3 or 10:2:6. since the order in B is not correct, we can eliminate II, and pick E as the correct answer. to verify III, suppose a=3, b=10, c=12 c^2= 144 b^2=100 a^2=9 a^2+b^2=109, which is less than c^2. so possible.



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M0831
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Updated on: 16 Jul 2016, 20:14
I think this is a highquality question and I agree with explanation.
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Originally posted by rishi02 on 16 Jul 2016, 07:22.
Last edited by rishi02 on 16 Jul 2016, 20:14, edited 1 time in total.



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Re M0831
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16 Jul 2016, 10:26
I think this is a highquality question and I agree with explanation.



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Re: M0831
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19 Feb 2017, 13:27
I was reading this post because I went wrong in this question, I am not sure that ration between angles is the reciprocal of the one of the sides since the sin theorem states that: A/sen(opposite)=B/sin(opposite)=C/sin(opposite) and sin is not a linear function



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Re: M0831
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22 Jul 2018, 01:26
Easy and very simple solution, thanks team!



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Re: M0831
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08 Nov 2018, 08:44
mvictor wrote: Bunuel 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 since we know the angles are in x, 3x, and 5x ratio, the sides must be in the same ratio. c:a:b=5:1:3 or 10:2:6. since the order in B is not correct, we can eliminate II, and pick E as the correct answer. I think that's wrong. Ratio of angles is not always the same as ratio of lengths, for example: Angles are in the ratio 1:2:3 > 306090 triangle lengths of 306090 triangle are in the ratio 1:\(\sqrt{3}\):2 and not 1:2:3



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Re: M0831
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10 Nov 2018, 02:27
Given sides a, b, and c to be the lengths of the three sides of triangle, with length c being the longest and a + b > c by the triangle inequality, then if the triangle is obtuse, then \(a^2 + b^2 < c^2\) and if the triangle is acute, then \(a^2 + b^2 > c^2\)
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