Remember to register **here** for FREE to ask any questions you may come across in your QCE studies!The year 12 syllabus forces you to do trigonometry in the context of non right-angled triangles as well. There's only two formulae to know: the cosine rule and the sine rule, both of which are conveniently on your formula sheet.

\begin{gather*}

c^2 = a^2+b^2-2ab\cos C \tag{cosine rule}\\

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\tag{sine rule}

\end{gather*}

Conventionally, the vertex \(A\) represents the angle \(\angle A\), and the side \(a\) is opposite to \(A\). This goes for the other variables as well.

The one question that I get asked all the time is this:

when do I use what?It's actually quite simple, provided you can follow this set of simple instructions. You start by thinking about what it is you're after.

________________________________________________

> If I am looking for a

**side** - If I am given one side and two angles, use the

**sine rule** - If I am given two sides and the included angle, use the

**cosine rule**. (Note: If you don't have the included angle, perhaps more work was necessary elsewhere)

> If I am looking for an

**angle** - If I am given all three sides, use the

**cosine rule** - If I am given two sides and an angle, use the

**sine rule**.

- Also, do not forget that angles in a triangle sum to \(180^\circ\).

________________________________________________

*Cosine rule to find an angle*This is generally the most straightforward case, because you'll know when you know all three sides of a triangle. The only thing to make sure is that you take \(C\) to be the side

**opposite** your angle.

For example, here the opposite side to the angle \(x\) has length \(3.8\).

\begin{align*}

3.8^2 &= 4.2^2 + 3.6^2 - 2\times4.2\times3.6 \cos x\\

30.24\cos x &= 16.16\\

x &= \cos^{-1} \frac{16.16}{30.24}\\

&\approx 57^\circ 42'

\end{align*}

*Cosine rule to find a side*Using the cosine rule to find a side is interesting. The following is the easy case, where the side we require is sandwiched between two sides of known lengths. It is very likely that it is the only case you will encounter.

We can just plug straight into the cosine rule here.

\begin{align*}

n^2 &= 11.8^2 + 9.3^2 - 2\times 11.8\times 9.3 \times \cos 101^\circ 45' \\

n &\approx \sqrt{270.42529}\\

&\approx 16.44

\end{align*}

*Sine rule to find a side*Two angles and a side is always enough to find the length of another side, using the sine rule. However the traps lie in that you may be given the wrong angles!

In the following example, suppose we wish to find the length of \(AC\). The problem lies in that we know the length \(BC = 12.7\), but we don't know the angle opposite it! That's the angle that we require for the sine rule, so we need to find it first.

But you may relax, because this is always doable. This is why I note that the angle sum of a triangle is \(180^\circ\). As a consequence, here we can deduce that \(\angle BAC = 180^\circ - 47^\circ -53^\circ = 80^\circ\).

*Now* we can throw everything into the sine rule. Note that the angle opposite to \(AC\) is \(\angle ABC\), with magnitude \(47^\circ\). Hence we consider:

\begin{align*}

\frac{AC}{\sin 47^\circ} &= \frac{12.7}{\sin 80^\circ}\\

AC &= \frac{12.7\sin 47^\circ}{\sin 80^\circ}\\

&\approx 9.43

\end{align*}

*Sine rule to find an angle* (NAIVELY)

Basically the same thing as above holds, from a naive perspective. Once again, to make life hard, I'll pick an example that leads to a wrong angle.

The given angle \(53.12^\circ\) is opposite a side of known length \(10.9\), which is good. The angle \(\alpha\) we require is not the one opposite the side of length \(8.7\), which is bad. So in fact, we need to find the other angle first. I'll label that angle \(\beta\).

\begin{align*}

\frac{10.9}{\sin 53^\circ 12'} &= \frac{8.7}{\sin \beta}\\

\sin \beta &= \frac{8.7 \sin 53^\circ 12'}{10.9}\\

\beta &\approx 39^\circ 44'.

\end{align*}

Then (surprise surprise...), we need the angle sum of a triangle to find \(\alpha\).

\[ \alpha \approx180^\circ - 53^\circ12' - 39^\circ 44' = 87^\circ 4'. \]

However, beware about this - this was mostly naive. There are some complications resulting from this method that have not been addressed in this post. To continue reading, visit the post regarding the "ambiguous case" of the sine rule.

An extra: Finding a side given two sides, but the corresponding angle is NOT an included angle.

In general, when we know two sides and any angle, we can always find the third side with the cosine rule. But when the angle is not sandwiched between the two known sides, but rather the unknown side helps cut it off, I found things get somewhat more bizarre.

Plugging into the cosine rule gives me

\begin{align*}

2^2 &= 4^2 + x^2 - 2\times 4\times x \times \cos 16^\circ\\

0 &= x^2 - (8\cos 16^\circ)x + 12

\end{align*}

...well then, it's a quadratic! And not a pleasant one.

\begin{align*}

x &= \frac{8\cos 16^\circ \pm \sqrt{64\cos^2 16^\circ - 48}}{2}\\

&\approx 5.51, \, 2.18

\end{align*}

What is going on here? Apparently there's TWO possible values for the length \(x\)?

This illustrates what's commonly known as the ambiguous case for the COsine rule. As this is not in your syllabus, we don't delve into it much here. If you're really interested, check out

this page I found.