# Triangle conditions

• Dec 26th 2010, 02:47 AM
sarahh
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?
• Dec 26th 2010, 03:11 AM
Quote:

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.
• Dec 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 :(
• Dec 26th 2010, 03:30 AM
Plato
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.
• Dec 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..
• Dec 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.
• Dec 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?
• Dec 26th 2010, 05:12 AM
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,

\$\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.