# Thread: Triangle conditions

1. ## Triangle conditions

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

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

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

2. Originally Posted by sarahh
In triangle ABC, if $\displaystyle a > b > c$ which of the following must be true?

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

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 $\displaystyle 60^o$

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

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

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

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

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

II. If $\displaystyle a=170^o$ and $\displaystyle c=1^o$ then $\displaystyle b=9^o$ contradicts the statement.

3. 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

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

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

But as Mr. Meade has shown, II can be false without contradiction.

5. 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..

6. 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.

7. 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?

8. 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,

$\displaystyle a=179^o$

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

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

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

Study Plato's responses as they are beautifully concise.