Ask Uncle Colin: A Troublesome Triangle

Dear Uncle Colin,

I couldn’t make head nor tail of this geometry problem: “If $a:b=12:7$, $c=3$, and $B\hat{A}C = 2 B\hat{C}A$, find the length of the sides $a$ and $b$.”

– Totally Rubbish In Geometry

Hi, TRIG, and thank you for your message! (And don’t put yourself down like that, it doesn’t help.)

I would probably start by letting $a=12x$ and $b=7x$, and seeing what came out of the equations. (I’d also let angle $B\hat{A}C=2\alpha$ and $B\hat{C}A=\alpha$, meaning $A\hat{B}C=\pi – 3\alpha$ – if I can avoid using that, I will.)

Let’s start with the sine rule

We have $\frac{\sin\br{\alpha}}{3}=\frac{\sin\br{2\alpha}}{12x}$

Multiplying around, that becomes $12x \sin(\alpha) = 3\sin(2\alpha)$, or $2x=\cos(\alpha)$ after using the double-angle formula. (For completeness: I’ve discarded the possibility that $\sin(\alpha)=0$ as that wouldn’t make for much of a triangle.)

How about the cosine rule?

We could probably bring in the other angle with the sine rule, but I don’t really fancy messing around with $\sin(3\alpha)$ right now.

Instead, let’s look at the cosine rule. I reckon $3^2 = (12x)^2 + (7x)^2 – 2(12x)(7x)\cos(\alpha)$.

Simplify that a bit to get $9 = 193x^2 – 168x^2 \cos(\alpha)$.

However, we know $\cos(\alpha)=2x$, so that’s $9 = 193x^2 – 336x^3$.

It turns out that $336x^3 – 193x^2 + 9$ factorises as $(16x+3)(7x-3)(3x-1)$, which gives $x = \frac{-3}{16}$, $x=\frac{3}{7}$ or $x=\frac{1}{3}$.

Three possibilities? Let’s check them!

Since $12x$ is a distance, the first solution is invalid.

The second solution gives $a=\frac{36}{7}$ and $b=3$, which seems superficially plausible but in fact doesn’t work: it gives an equilateral triangle, with the apex double the size of the two base angles. However, the only triangle that fits that description is (don’t tell the Ninja) the 90-45-45 triangle. This triangle clearly isn’t that!1

So we’re left with the last one, with $a=4$ and $b=\frac{7}{3}$.

Hope that helps!

– Uncle Colin


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. What’s happened here? I think the apex angle is very large, and is such that its sine is equal to the sine of double each base angle – but the curse of many-to-one functions is that that doesn’t mean the apex angle is double the base angle! []


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter