Dear Uncle Colin,

I have a triangle with sides 4.35cm, 8cm and 12cm; the angle opposite the 4.35cm side is 10º ((Degrees? Yeah, I know. If the question’s in degrees, we work in degrees, however silly they are. Same goes for furlongs per fortnight.)) and need to find the largest angle.

I know how to work this out in two ways: I can use the cosine rule with the three sides, which gives me 151.3º – I believe that to be correct – or I can use the sine rule, which gives me 28.6º. Why am I getting two different answers?

-- Surely It’s Not Evident

Dear SINE,

Ah! You have stumbled onto one of the most frustrating things about trigonometry: that the functions are not one-to-one. That means, there are many different angles that have the same sine, cosine or tangent (in fact, infinitely many). For example, $\cos(45º) = \cos(315º)$ – check on your calculator if you like; they’re both $\frac{\sqrt{2}}{2}$.

When you’re dealing with triangles, though, you’re usually shielded from this – except in one specific setup, when you’re finding an angle using the sine rule (and even then, “more triangles” in some sense are acute than obtuse.)

So, you’ve got a bit unlucky with this one!

You can immediately say that the largest angle is opposite the longest side, and can write down $\frac{\sin(x)}{12} = \frac{\sin(10º)}{4.35}$.

Multiplying up gives $\sin(x) = \frac{12 \sin(10º)}{4.35} \approx 0.479$.

Your instinct is to hit the $\sin^{-1}$ button, which gives you 28.6º – only a moment’s thought says that can’t be the biggest angle if they’re going to add up to 180º.

Instead, you need to use the fact that $\sin(x) = \sin(180º - x)$ and say $x = 151.4º$ (if you round it carefully).

The reason for that fact lies in the symmetry of the sine graph – but I’ll save that for another day!

-- Uncle Colin