In the triangle shown, what is the least possible value of for \(\angle A\) to be the largest angle of the triangle, it must be that \(m < x < n\) . What is the least possible value of \(n-m\), expressed as a common fraction?

Guest Oct 6, 2018

edited by
Guest
Oct 6, 2018

#1**+1 **

This is what I found.

5/3 < x < 13

As x tends to 5/3 angle A tends to 180 degrees

As x tends to 13 angle A tends to 0 degrees

so 0 < angle A < 180

m=0, n=180

n-m=180

Melody
Oct 7, 2018

#3**+1 **

Yea ok, Now I re-read it I I am not so sure what the question is asking for.

How can the least possible value of something be between two values. Wouldn't if just be the smallest of the two?

Maybe if I play with it again I might work out what on Earth the question actually is.

The largest value for x

Melody
Oct 7, 2018

#4**+1 **

I think he needs to find the lower and upper bounds of all the values of x for which angle A in the triangle is the largest angle, but I'm not sure

Guest Oct 7, 2018

#6**0 **

Hi, I'm the person who originally asked the question. I have personally been stuck on this problem for a very long time, so I'm severely at a loss and cannot properly provide you extra information. I'm pretty certain that you do not need to find \(x\) and should instead focus on finding the least possible value of \(x\) which is \(m\) and the greatest possible value of \(x\) which is \(n\). Thanks and sorry if this doesn't help!

Guest Oct 10, 2018

#5**+1 **

If A is to be the largest angle then BC must be the largest side,

so x + 9 > 3x,

x < 9/2.

x = 4.5 returns a value for A of 71.65 deg, ( approximately ).

That will also be the value of angle B, so for a unique smallest/largest angle I suppose we just add a tiddly bit to the 71.65.

Without knowing what m and n are supposed to be, I don't see how we can answer the last part of the question.

Guest Oct 7, 2018