The sine rule

In any triangle, a side and the angle opposite it form a pair. The sine rule says the ratio of a side to the sine of its opposite angle is the same for all three pairs:

The letters match up: side $a$ is opposite angle $A$, and so on.

This is stated in most courses as something to use. Here is why it is true.

Before you start

You need two things:

• The sine ratio. In a right-angled triangle, $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$.
• How to rearrange a formula. If $b\sin A = a\sin B$ then dividing both sides by $\sin A \sin B$ gives $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$.

Predict first

Below is a triangle with a line dropped from the top corner straight down to the base, at right angles.

Before you move anything, predict: as the top corner slides along the top, the height of that line changes. Which of the two sides — $a$ or $b$ — grows when the angle opposite it grows?

Write your prediction down, then test it. Watch what happens to the height.

The proof

Take a triangle with corners $A$, $B$ and $C$. Write the side opposite $A$ as $a$, opposite $B$ as $b$, and opposite $C$ as $c$.

Drop a perpendicular. From $C$, draw a line straight down to the side $AB$, making a right angle where it lands. Call the place it lands $D$, and call the length $CD$ the height $h$.

That line splits the triangle into two right-angled triangles, $ACD$ and $BCD$. Both of them contain the same $h$, and that is the whole proof: two ways of writing one length.

In the left-hand triangle, $ACD$:

• the hypotenuse is $AC$, which has length $b$;
• the side opposite angle $A$ is $CD$, which has length $h$.

So the sine ratio gives

In the right-hand triangle, $BCD$:

• the hypotenuse is $BC$, which has length $a$;
• the side opposite angle $B$ is again $CD$, with length $h$.

So

Equate the two. Both expressions equal the same height, so they equal each other:

Divide both sides by $\sin A \sin B$:

That is two thirds of the sine rule. For the third ratio, do the same thing from a different corner — drop the perpendicular from $B$ instead of $C$ — and you get $\dfrac{c}{\sin C}$ equal to the other two. Nothing in the argument used the shape of the triangle, so it holds for every triangle.

Why the perpendicular is allowed to land where it does

In the triangle above, $D$ sits between $A$ and $B$. In an obtuse triangle it can land outside the side, and the proof needs one extra fact: $\sin(180^\circ - \theta) = \sin\theta$. An obtuse angle and its supplement have the same sine, so the angle you measure inside the right-angled triangle gives the same sine as the triangle's own angle, and the algebra is unchanged.

That is worth knowing rather than skipping, because a proof that only works for acute triangles would not be a proof of the sine rule.

Check your understanding

1. Use the rule to write down the ratio for the third pair, given $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.
2. In a triangle, $a = 9$, $A = 40^\circ$ and $B = 65^\circ$. Which ratio would you use to find $b$?
3. Explain why the proof needs two right-angled triangles rather than one.