# Homework Help: Triangle length and largest angle

1. May 1, 2012

### 314Jason

1. The problem statement, all variables and given/known data

A triangle has sides of length (n2+n+1), (2n+1) and (n2-1), where n > 1.

(a) Explain why the side (n2+n+1) must be the longest side of the triangle

(b) Show that the largest angle, θ , of the triangle is 120º.

2. Relevant equations

In a triangle of sides a, b and c: a - b < c < a + b.

3. The attempt at a solution

(a)

n2+n+1 > 2n+1
n2-n > 0
n(n-1) > 0
Thus, n < 0, n > 1.

n2+n+1 > n2-1
n+1 > -1
n > -2

I thought they would both give n > 1. I don't know what other way to show this is true.

(b) I have no idea what to do here. I thought about vectors, but don't know where to go from there.

2. May 1, 2012

### 314Jason

I got (b)! I can just public the sides into the cos rule and rearrange to find theta. The n should cancle out.

Does anyone know how to do (a)? Please?

3. May 1, 2012

### Curious3141

First of all, it should be immediately obvious that for n>1,

(n2+n+1) > (n2-1)

Now all you need to prove is that for n>1,

(n2+n+1) > (2n+1)

Consider the opposite proposition: (n2+n+1) $\leq$ (2n+1)

and solve the inequality. What's the possible range for n you get?

4. May 1, 2012

### tiny-tim

Hi 314Jason!
but you have proved it …

you've proved it's true if {n < 0 or n > 1} and {n > -2} …

so, in particular, it's true for {n > 1} !

5. May 1, 2012

### Curious3141

This is what comes of not reading the OP thoroughly.

6. May 1, 2012

### 314Jason

So because n > 1 is in both equations, this is true?

7. May 1, 2012

### tiny-tim

yes, the question specifies that n > 1, so you only have to prove it for n > 1

8. May 1, 2012

### 314Jason

Ok then, thanks!