# Ask Uncle Colin: A Tricksy Triangle

Dear Uncle Colin,

I’m given a triangle ABC such that angle $B$ is $xº$, angle $C$ is $3xº$, and the sides are labelled as you’d expect. I have to prove that $a = 4b \cos(2x)\cos(x)$, but I’m going round in circles! What do you recommend?

Triangle Rules: It’s Common Knowledge? Says You!

Hi, TRICKSY, and thanks for your message!

As usual, I’m going to try exploring two methods. In this case, I’ve got one that works and one that… might work.

### One that works

Angle $A$ is clearly $(180 - 4x)º$, so $\sin(A) = \sin(4x)$.

Taking the degrees as read, that means $\frac{a}{\sin(4x)} = \frac{b}{\sin(x)}$, or $a = b\frac{\sin(4x)}{\sin(x)}$.

What’s $\sin(4x)$? It’s $\sin(2(2x))$, or $2\sin(2x)\cos(2x)$ – or even $4\sin(x)\cos(x)\cos(2x)$, which looks handy!

Substituting that in gives $a = 4b\cos(x)\cos(2x)$, as required!

### One that might

Another option is to use the cosine and sine rules to find and eliminate side $c$. I’m skeptical, but let’s try.

- Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos(3x)$
- Sine rule: $\frac{c}{\sin(3x)} = \frac{b}{\sin(x)}$

So $c = b\frac{\sin(3x)}{\sin(x)}$

Now, $\sin(3x) = 3\sin(x) - 4\sin^2(x)$ (after a bit of expansion), so $c = b\br{3 - 4\sin^2(x)}$, or $b\br{4\cos^2(x) - 1}$.

What about $\cos(3x)$? That’s $4\cos^3(x) - 3\cos(x)$

OK. So, letting $K = \cos(x)$ the cosine rule becomes:

- $b^2 \br{4K^2 - 1}^2 = a^2 + b^2 - 2ab\br{4K^3 - 3K}$, or
- $b^2 \br{16K^4 - 8K^2} = a^2 - 2ab\br{4K^3 - 3K}$, or
- $a^2 - 2ab\br{4K^3 - 3K} - b^2 \br{16K^4 - 8K^2}$.

Where are we trying to get, again? $a = 4b\cos(2x)\cos(x)$. Can we rewrite that to see if there’s any help to be had? It’s $\frac{a}{b} = 4 K\br{2K^2 - 1}$.

Let’s divide everything by $b^2$ and let $\lambda = \frac{a}{b}$:

- $\lambda^2 - 2\lambda\br{4K^3 - 3K} - \br{16K^4 - 8K^2} = 0$

Complete the square:

- $\br{\lambda - (4K^3 - 3K)}^2 - \br{4K^3 - 3K}^2 = 16K^4 - 8K^2$
- $\br{\lambda - (4K^3 - 3K)}^2 = \br{16K^6 -24K^4 + 9K^2} + 16K^4 - 8K^2$
- $\dots = 16K^6 -8K^4 + K^2$
- $\dots = K^2 \br{4K^2 - 1}^2$

So:

- $\lambda = \br{4K^3 - 3K} \pm \br{4K^3 - K}$

This gives either $-2K$ (which is no good, as it’s a ratio of lengths) or $8K^3 - 4K$, which is $4K\br{2K^2 - 1}$, as required.

Urghf. I prefer the first way.

Hope that helps!

- Uncle Colin