Triangle conditions

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December 26th 2010, 02:47 AM
sarahh

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Triangle conditions

In triangle ABC, if $a > b > c$ which of the following must be true?

I. $60 < a < 180$
II. $45 < b < 90$
III. $0 < c < 60$

Why is it only I and III? How can I both prove these are correct and disprove II?
December 26th 2010, 03:11 AM
Archie Meade

Quote:

Originally Posted by sarahh

In triangle ABC, if $a > b > c$ which of the following must be true?

I. $60 < a < 180$
II. $45 < b < 90$
III. $0 < c < 60$

Why is it only I and III? How can I both prove these are correct and disprove II?

I. If the triangle was equilateral, then all 3 angles would be $60^o$

Hence, if the triangle has a largest angle, it is greater than $60^o$

while all angles must be less than $180^o$ so that they can sum to $180^o$

III. If one angle is greater than $60^o$ than another must be less than $60^o.$

This is because the two remaining angles sum to less than $120^o$

and if they were equal, they'd both be less than $60^o.$

II. If $a=170^o$ and $c=1^o$ then $b=9^o$ contradicts the statement.
December 26th 2010, 03:16 AM
sarahh

Hmm but how come that would be 2's explanation (the triangle inquality theorem would fail wouldn't it?) Also, I still don't understand why angle a could be 179 (since that value would fit the bounds given--again the triangle inequality theorem would fail for angles b and c.) Maybe if we took some concrete examples? I still don't see it yet :(
December 26th 2010, 03:30 AM
Plato

The key is in the given, $a>b>c$ .
If I were false the all three angles are less than $60$ : impossible.

If III were false the all three angles are greater than $60$ : impossible.

But as Mr. Meade has shown, II can be false without contradiction.
December 26th 2010, 03:45 AM
sarahh

But wait, that is just one way to read part I..if a = 179 how could this be? What about an example of: a = 80, b = 70, and c = 30, where all 3 are true above..
December 26th 2010, 04:08 AM
Plato

Quote:

Originally Posted by sarahh

But wait, that is just one way to read part I..if a = 179 how could this be? What about an example of: a = 80, b = 70, and c = 30, where all 3 are true above..

All of that is true. But it has nothing to do with the question.
The question ask: Which of these conditions must be true?
You have been shown that I & III must be true but II can be false.
December 26th 2010, 04:44 AM
sarahh

That's kinda my point too, I mean if it has to be true for all possible numbers in the interval then why could a = 179 in the first case?
December 26th 2010, 05:12 AM
Archie Meade

Quote:

Originally Posted by sarahh

That's kinda my point too, I mean if it has to be true for all possible numbers in the interval then why could a = 179 in the first case?

Hi sarahh,

$a=179^o$

$b+c=1^o,\;\;b>c$

I and III are vindicated as normal but II is violated.

$b=0.8^o,\;\;c=0.2^o$ is one of multiple ways to combine b and c.

Study Plato's responses as they are beautifully concise.